9.2. math — Mathematical functions — Python 3.6.5 documentation
— Mathematical functions
This module is always available.
It provides access to the mathematical
functions defined by the C standard.
These functions cannot be used
use the functions of the
same name from the
module if you require support for complex
The distinction between functions which support complex numbers and
those which don’t is made since most users do not want to learn quite as much
mathematics as required to understand complex numbers.
Receiving an exception
instead of a complex result allows earlier detection of the unexpected complex
number used as a parameter, so that the programmer can determine how and why it
was generated in the first place.
The following functions are provided by this module.
Except when explicitly
noted otherwise, all return values are floats.
9.2.1. Number-theoretic and representation functions
math.ceil(x)
Return the ceiling of x, the smallest integer greater than or equal to x.
If x is not a float, delegates to x.__ceil__(), which should return an
math.copysign(x, y)
Return a float with the magnitude (absolute value) of x but the sign of
On platforms that support signed zeros, copysign(1.0, -0.0)
returns -1.0.
math.fabs(x)
Return the absolute value of x.
math.factorial(x)
Return x factorial.
if x is not integral or
is negative.
math.floor(x)
Return the floor of x, the largest integer less than or equal to x.
If x is not a float, delegates to x.__floor__(), which should return an
math.fmod(x, y)
Return fmod(x, y), as defined by the platform C library. Note that the
Python expression x % y may not return the same result.
The intent of the C
standard is that fmod(x, y) be exactly ( to infinite
precision) equal to x - n*y for some integer n such that the result has
the same sign as x and magnitude less than abs(y).
Python’s x % y
returns a result with the sign of y instead, and may not be exactly computable
for float arguments. For example, fmod(-1e-100, 1e100) is -1e-100, but
the result of Python’s -1e-100 % 1e100 is 1e100-1e-100, which cannot be
represented exactly as a float, and rounds to the surprising 1e100.
this reason, function
is generally preferred when working with
floats, while Python’s x % y is preferred when working with integers.
math.frexp(x)
Return the mantissa and exponent of x as the pair (m, e).
m is a float
and e is an integer such that x == m * 2**e exactly. If x is zero,
returns (0.0, 0), otherwise 0.5 &= abs(m) & 1.
This is used to “pick
apart” the internal representation of a float in a portable way.
math.fsum(iterable)
Return an accurate floating point sum of values in the iterable.
loss of precision by tracking multiple intermediate partial sums:
&&& sum([.1, .1, .1, .1, .1, .1, .1, .1, .1, .1])
&&& fsum([.1, .1, .1, .1, .1, .1, .1, .1, .1, .1])
The algorithm’s accuracy depends on IEEE-754 arithmetic guarantees and the
typical case where the rounding mode is half-even.
On some non-Windows
builds, the underlying C library uses extended precision addition and may
occasionally double-round an intermediate sum causing it to be off in its
least significant bit.
For further discussion and two alternative approaches, see the .
math.gcd(a, b)
Return the greatest common divisor of the integers a and b.
a or b is nonzero, then the value of gcd(a, b) is the largest
positive integer that divides both a and b.
gcd(0, 0) returns
New in version 3.5.
math.isclose(a, b, *, rel_tol=1e-09, abs_tol=0.0)
Return True if the values a and b are close to each other and
False otherwise.
Whether or not two values are considered close is determined according to
given absolute and relative tolerances.
rel_tol is the relative tolerance – it is the maximum allowed difference
between a and b, relative to the larger absolute value of a or b.
For example, to set a tolerance of 5%, pass rel_tol=0.05.
The default
tolerance is 1e-09, which assures that the two values are the same
within about 9 decimal digits.
rel_tol must be greater than zero.
abs_tol is the minimum absolute tolerance – useful for comparisons near
zero. abs_tol must be at least zero.
If no errors occur, the result will be:
abs(a-b) &= max(rel_tol * max(abs(a), abs(b)), abs_tol).
The IEEE 754 special values of NaN, inf, and -inf will be
handled according to IEEE rules.
Specifically, NaN is not considered
close to any other value, including NaN.
inf and -inf are only
considered close to themselves.
New in version 3.5.
– A function for testing approximate equality
math.isfinite(x)
Return True if x is neither an infinity nor a NaN, and
False otherwise.
(Note that 0.0 is considered finite.)
New in version 3.2.
math.isinf(x)
Return True if x is a positive or negative infinity, and
False otherwise.
math.isnan(x)
Return True if x is a NaN (not a number), and False otherwise.
math.ldexp(x, i)
Return x * (2**i).
This is essentially the inverse of function
math.modf(x)
Return the fractional and integer parts of x.
Both results carry the sign
of x and are floats.
math.trunc(x)
Return the
value x truncated to an
(usually an integer). Delegates to
have a different call/return pattern
than their C equivalents: they take a single argument and return a pair of
values, rather than returning their second return value through an ‘output
parameter’ (there is no such thing in Python).
For the , , and
functions, note that all
floating-point numbers of sufficiently large magnitude are exact integers.
Python floats typically carry no more than 53 bits of precision (the same as the
platform C double type), in which case any float x with abs(x) &= 2**52
necessarily has no fractional bits.
9.2.2. Power and logarithmic functions
math.exp(x)
Return e**x.
math.expm1(x)
Return e**x - 1.
For small floats x, the subtraction in exp(x) - 1
function provides a way to compute this quantity to full precision:
&&& from math import exp, expm1
&&& exp(1e-5) - 1
# gives result accurate to 11 places
1.9649e-05
&&& expm1(1e-5)
# result accurate to full precision
1.6668e-05
New in version 3.2.
math.log(x[, base])
With one argument, return the natural logarithm of x (to base e).
With two arguments, return the logarithm of x to the given base,
calculated as log(x)/log(base).
math.log1p(x)
Return the natural logarithm of 1+x (base e). The
result is calculated in a way which is accurate for x near zero.
math.log2(x)
Return the base-2 logarithm of x. This is usually more accurate than
log(x, 2).
New in version 3.3.
returns the number of bits necessary to represent
an integer in binary, excluding the sign and leading zeros.
math.log10(x)
Return the base-10 logarithm of x.
This is usually more accurate
than log(x, 10).
math.pow(x, y)
Return x raised to the power y.
Exceptional cases follow
Annex ‘F’ of the C99 standard as far as possible.
In particular,
pow(1.0, x) and pow(x, 0.0) always return 1.0, even
when x is a zero or a NaN.
If both x and y are finite,
x is negative, and y is not an integer then pow(x, y)
is undefined, and raises .
Unlike the built-in ** operator,
converts both
its arguments to type .
Use ** or the built-in
function for computing exact integer powers.
math.sqrt(x)
Return the square root of x.
9.2.3. Trigonometric functions
math.acos(x)
Return the arc cosine of x, in radians.
math.asin(x)
Return the arc sine of x, in radians.
math.atan(x)
Return the arc tangent of x, in radians.
math.atan2(y, x)
Return atan(y / x), in radians. The result is between -pi and pi.
The vector in the plane from the origin to point (x, y) makes this angle
with the positive X axis. The point of
is that the signs of both
inputs are known to it, so it can compute the correct quadrant for the angle.
For example, atan(1) and atan2(1, 1) are both pi/4, but atan2(-1,
-1) is -3*pi/4.
math.cos(x)
Return the cosine of x radians.
math.hypot(x, y)
Return the Euclidean norm, sqrt(x*x + y*y). This is the length of the vector
from the origin to point (x, y).
math.sin(x)
Return the sine of x radians.
math.tan(x)
Return the tangent of x radians.
9.2.4. Angular conversion
math.degrees(x)
Convert angle x from radians to degrees.
math.radians(x)
Convert angle x from degrees to radians.
9.2.5. Hyperbolic functions
are analogs of trigonometric functions that are based on hyperbolas
instead of circles.
math.acosh(x)
Return the inverse hyperbolic cosine of x.
math.asinh(x)
Return the inverse hyperbolic sine of x.
math.atanh(x)
Return the inverse hyperbolic tangent of x.
math.cosh(x)
Return the hyperbolic cosine of x.
math.sinh(x)
Return the hyperbolic sine of x.
math.tanh(x)
Return the hyperbolic tangent of x.
9.2.6. Special functions
math.erf(x)
Return the
function can be used to compute traditional statistical
functions such as the :
def phi(x):
'Cumulative distribution function for the standard normal distribution'
return (1.0 + erf(x / sqrt(2.0))) / 2.0
New in version 3.2.
math.erfc(x)
Return the complementary error function at x.
is defined as
1.0 - erf(x).
It is used for large values of x where a subtraction
from one would cause a .
New in version 3.2.
math.gamma(x)
Return the
New in version 3.2.
math.lgamma(x)
Return the natural logarithm of the absolute value of the Gamma
function at x.
New in version 3.2.
9.2.7. Constants
The mathematical constant π = 3.141592…, to available precision.
The mathematical constant e = 2.718281…, to available precision.
The mathematical constant τ = 6.283185…, to available precision.
Tau is a circle constant equal to 2π, the ratio of a circle’s circumference to
its radius. To learn more about Tau, check out Vi Hart’s video , and start celebrating
by eating twice as much pie!
New in version 3.6.
A floating-point positive infinity.
(For negative infinity, use
-math.inf.)
Equivalent to the output of float('inf').
New in version 3.5.
A floating-point “not a number” (NaN) value.
Equivalent to the output of
float('nan').
New in version 3.5.
CPython implementation detail: The
module consists mostly of thin wrappers around the platform C
math library functions.
Behavior in exceptional cases follows Annex F of
the C99 standard where appropriate.
The current implementation will raise
for invalid operations like sqrt(-1.0) or log(0.0)
(where C99 Annex F recommends signaling invalid operation or divide-by-zero),
for results that overflow (for example,
exp(1000.0)).
A NaN will not be returned from any of the functions
above unless one or more of the input arguments was a NaN; in that case,
most functions will return a NaN, but (again following C99 Annex F) there
are some exceptions to this rule, for example pow(float('nan'), 0.0) or
hypot(float('nan'), float('inf')).
Note that Python makes no effort to distinguish signaling NaNs from
quiet NaNs, and behavior for signaling NaNs remains unspecified.
Typical behavior is to treat all NaNs as though they were quiet.
Complex number versions of many of these functions.1=5,2=15,3=75.4=1250.5=?
1=5,2=15,3=75.4=1250.5=?
1、“脑筋急转弯+逻辑推理”：00001因为已知5=1,所以1=5；因为每个等式左边数同时还代表了右边结果的位数,所以推断所求数字必为5位数）2、“谐音”：5=219451个字 5（我）2个字 15（你我）3个字 215（爱你我）4个字 2145（爱你是我） 5个字 21945（爱你就是我）3、“数字推理”：5是公有的,因此不必考虑.直接看5前面的数字（即0,1,21,214,),相加（分别得0,1,3,7,),再相减（即得1,2,4）.如下：1 5 0 1-0=1 1为2的0次方2 15 1 3-1=2 2为2的1次方 3 215 2+1=37-3=4 4为2的2次方4 +4+5=7x-7=8 不完全归纳法推断应为2的3次方即8,5 214y5 x=15 故 y=15-7=8所以 5=21485
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与《1=5,2=15,3=75.4=1250.5=?》相关的作业问题
56*15+75*85 + 29*15=15*(56+29) + 75*85=15*85 + 75*85=85*(15+75)=85*100=85*(100-10)=8500 - 850= 7650
50——fifty30——thirty15——fifteen5——five75——seventy five20——twenty90——ninety
利用公式sin(x+y)=sinxcosy+cosysinxcos(x+y)=cosxcosy-sinxsiny
白线为尺子,红色为法线（尺子是有量度的）；能画出15度和75度,105度和165度就很容易解了.注：尺子是30度,60度和90度的.希望你能看明白,其实也不难
15=45-而45 和30的参数是已知的
1.w=4 f(15')= f(75')=1/2 2.最大值为23.a+b=0
sin15=cos75=sin165=-cos15 这样说吧 两个角相加=90度,则一个角的COS值=另一个角的Sin值 ； 若相加=180度,则两角Sin值相等,COS值互为相反数
15和75的最小公倍数是最大公约数的(5)倍；它们的最大公约数是最小公倍数的(1/5),这两个数的最小公倍数是75
7.25-0.87-2.13=7.25-（0.87+2.13）=7.25-3=4.==846×99+46=46×（99+1）=46×100=-6.75+2.08=9.15+2.08=11.23
1 12 4 1 15 13
15/0.75+15/0.25=15÷3/4+15÷1/4=15×4/3+15×4=5×4+15×4=（5+15）×4=20×4=80
SIN 15=0.. , 180=0 , 270=-1 , 360=0COS 15=0..2588 , 90=0 , 180=-1 , 270=0 , 360=1TAN 15=0.2679 , 75=3.732 , 90=不存在 , 180=0 ,270=不存
15和75绝对没错的~
4分之(3 )=15÷( 20)=0.75=(32 )分之24 再问： 3/4=15/20=18/24 这个是对的吗？ 再答： 对的
720×（400÷800）＝360 60÷（120÷30）＝15 3.75×（5/8×8/5）＝3.75 6÷（4/3÷4/3）＝6/1
用电脑答起来有点麻烦,所以多给点分吧:)因为∠1=∠2所以AB‖EF所以∠MAE=∠AEF=45°∠AEG=∠AEF+=∠FEG=45°+15°=60°因为EG平分∠AEC,所以∠GEC=∠AEG=60°所以∠FEC=∠FEG+∠GEC=15°+60°=75°所以∠FEC=∠NCE所以EF‖CD因为AB‖EF,EF‖C
三角尺有90°,60°,30°,45°画出15°先在纸上先画出45°角以这角一边为公共边向内部画个30°的角75°可以认为是45°+30°105°=15°+90°135°=90°+45°150°=90°+60°设X和Y是随机变量,且有E(X)=3,E(Y)=1,D(X)=4,D(Y)=9 .令Z=5X-Y+15,已知X与Y的相关
设X和Y是随机变量,且有E(X)=3,E(Y)=1,D(X)=4,D(Y)=9 .令Z=5X-Y+15,已知X与Y的相关系数是0.25,求D(X)
cov(x,y)=2*3/4=3/2D(z)=25D(x)+D(Y)+2cov(5x,y)=136+10cov(x,y)=151如有意见,欢迎讨论,共同学习；如有帮助,
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与《设X和Y是随机变量,且有E(X)=3,E(Y)=1,D(X)=4,D(Y)=9 .令Z=5X-Y+15,已知X与Y的相关》相关的作业问题
E(Z)=5E(X)-E(Y) 15=-1D(Z)=D(5X-Y)=25D(X) D(Y)=109
1)x12=[-1±√(1²-4*1*4)]/2=-1/2±√15i/2∴x1=-1/2+√15i/2 x2=-1/2-√15i/22)设z=a+bi ∵a+bi-2√(a²+b²)=-7+4i ∴b=4 a-2√(a²+16)=-7 => 4a²+64=a²
因为M(X):M(Y)=6:1,所以在Z=X+2Y的反应中,m(X):m(Y)=6:(1*2)=3:1.所以产物中x的质量占总质量的3/4即m(X)=16g*3/4=12g.
1.设A（x1,x1+1) B(x2,X2+1) ,C为AB中点,则向量OC与直线AB垂直,将直线带入方程中,化简得到,(a^2+b^2)X^2 + 2a^2X + a^2-a^2b^2=0.通过韦达定理可以写出C点坐标.将向量OC与向量（1,1） 相乘为零可得 a=
img class="ikqb_img" src="http://a.hiphotos.baidu.com/zhidao/wh%3D600%2C800/sign=fa954f8fb27ecaa113bbe5/7af40ad162d9f2d31fa8cd43abec8a136227cce5.jpg"
已知m,x,y满足1.2/3(x-5)^2+5|m|=0 x-5=0 m=0x=5 m=02.-2a^2b^y+1与7a^3b^2是同类项.2+y+1=3+2y=2代数式2x^2-6y^2+m(xy-9y^2)-(3x^2-3xy+7y^2)的值.=2x^2-6y^2+0*m(xy-9y^2)-3x^2+3xy-7y^
反比例函数Y=3M-1/X的图象在第一.三象限内,说明3M-1>0;一次函数Y=（1-M）X+2的图限经过一、二、三象限,说明1-M>0；解这两个不等式得到解集：1/3
1.k为三2.k大于13.x1小于x2
告诉你会害了你有类似的题目,自己好好想想已知1/x-1/y=3,求分式2x+3xy-2y/x-2xy-y的值根据条件可得y-x=3xy(方程1/x-1/y=3两边同乘xy) 所以分式的分子可以化简为2x+3xy-2y =2（x-y）+3xy =-6xy+3xy =-3xy 分式的分母x-2xy-y =-3xy-2xy=
已知反比例函数y=(1-2k)/x,当x＞0时,y随x的增大而减小,则k的取值范围为（k＜1/2）.这时图像经过（一三）象限.1-2k＞0;k＜1/2;很高兴为您解答,skyhunter002为您答疑解惑如果本题有什么不明白可以追问, 再问： y=(1-2k)/x,当x＞0时，y随x的增大而减小， 怎么确定k的取值范围
图像过A(2,m),m=k/2,k=2mB(2,0),△AOB的面积是1/2（1/2)*2*m=1/2,m=1/2 k=1反比例函数:y=1/x根据对称性,|PQ|=2|OP|P(x0,y0),y0=1/x0|OP|^2=x0^2+y0^2=x0^2+1/x0^2≥2√(x0^2/x0^2)=2|OP|≥√2所以|PQ
把B点坐标代入反比例函数的y=2/x,2n=2,n=1,再把A(2,1) B(-1,-2)代入一次函数求出k2=1,b=-1即y=x-1△APO∽△AOB,AP/OA=OA/AB,AP=5倍根号2/6,P在直线AB上y=x-1再利用两点间距离公式可求出点坐标,P（7/6,1/6)
1、x=2,y=0,所以4x^2-2xy+2y^2=4*4-0+0=162、x+3=5-y,所以x+y=2；a,b互为倒数,所以ab=1所以0.5(x+y）+5ab=0.5*2+5*1=6
(3a+2)x²+(5b-3)xy-(9a+10b)y²-x+2y-6不含二次项,∴3a+2=05b-3=09a+10b=03a=-25b=33a+5b=-2+3=1
已知X=2,Y=3是3X-4MY+9=0的一个解,则有：3*2-4m*3+9=0即2 - 4m+3=04m=5解得：m=4分之5
圆心坐标M（2,0）已知圆x²+y²-4x-5＝0的弦AB中点P(3,1)则 MP⊥ABkMP=1 kAB=-1点斜式：y-1=-(x-3)整理得一般式：x+y-4=0AB直线方程为 x+y-4=0
题目比较简单,概念问题,要把相关定义搞清楚.B点坐标(-4,0) C点在L1上,所以C(-4,6) 因为ABCD为矩形 所以CD||BA 从而D点纵坐标为6,而D在L2上,所以D点坐标（-1,6）F点坐标：解方程 y=2-x y=2x+8 所以x=-2,y=4 所以F点坐标（-2,4）长度：DC=AB=-1-（-4）=
(1)已知x^2+y^2=5 [1]xy=2 [2][1]+[2]*2 (x+y)^2=9x+y=±3所以1/(1/x+1/y)＝xy/(x+y)=2/(±3)=±2/3(2)计算:1/(a-x)-1/(a+x)-2x/(a^2+x^2)-(4x^3)/(a^4+x^4)+(8x^7)/(x^8-a^8).=(a+x-已知tana/1-tana=1,求下列各式的值：（1）sina-cosa/sina+cosa：（2）sin^2a+sin
已知tana/1-tana=1,求下列各式的值：（1）sina-cosa/sina+cosa：（2）sin^2a+sinacosa+2：...已知tana/1-tana=1,求下列各式的值：（1）sina-cosa/sina+cosa：（2）sin^2a+sinacosa+2：
tana=1-tana则sina/cosa=tana=1/2cosa=2sinasin²a+cos²a=1所以sin²a=1/5sinacosa=2sin²a=2/5所以(1)原式=(sina-2sina)/(sina+2sina)=-1/3(2)原式=1/5+2/5+2=13/5
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与《已知tana/1-tana=1,求下列各式的值：（1）sina-cosa/sina+cosa：（2）sin^2a+sin》相关的作业问题
log3(4)=lg4/lg3=2lg2/lg3=2a/blog2(12)=lg12/lg2=(lg3+2lg2)/lg2=(2a+b)/a
sinx-cosx=1/5sinx ²+cosx ²-2sinx cosx =(sinx-cosx)²=1/25sinxcosx=12/25sin^3x-cos^3x=(sinx-cosx)(sin^2x +cos^2x+sinxcosx)=(sinx-cosx)(1+12/25)=1/5
∵a-b=5 ab=4 ∴1.3a²+3b²=3(a²+b²)3[(a²-2ab+b²)+2ab]=3[(a-b)²+2ab]=3×(5²+8)=992.(a+b)²=a²+2ab+b²=a²-2ab
lg6=lg(2×3）=lg2+lg3=a+blog(3)4=lg4/lg3=(2lg2)/lg3=(2a)/blog(2)12=lg12/lg2=(lg4×3)/lg2=(lg4+lg3)/lg2=(2lg2+lg3)/lg2=(2a+b)/alg(2/3)=lg2-lg3=a-
（1）lg6=lg2+lg3=a+b （lgab=lga+lgb）(2)log3|4=lg4/lg3=2lg2/lg3=2a/b (loga|b=lgb/lga lga^n=nlga)(3)log2|12=lg12/lg2=(lg4+lg3)/lg2=(2a+b)/a=2+b/a(4)lg3/2=lg3-lg2=b-a
lg6=lg2X3=a+b,log34=用换底公式=2a除b,第三题还用换底公式=b+2a除以a,第四题b-a 再问： 呃 …恕我愚钝 第三题还是不懂 再答： 换底公式，lg2x2x3除以lg2，等于lg2+lg2+lg3除以lg2，2a+b/a
tanα/(tanα-1)=-1∴tanα=1/2sin²α+sinαcosα+2=（sin²α+sinαcosα+2sin²α+2cos²α)/(sin²α+cos²α)=(3sin²α+2cos²α+sinαcosα)/(sin&sup
=-5/3tanα/(tanα-1)=1,tanα=1/2sinα=1/2cosα∴(sinα-3cosα)/(sinα+cosα)=(1/2cosα-3cosα)/(1/2cosα+cosα)=(-5/2)/(3/2)=-5/3
（1）lg6=lg2x3=lg2+lg3=a+b（2)log3|4=lg3-lg4=lg3-2lg2=b-2a(3)log2|12=lg1/6=-lg6=-lg2-lg3=-a-b(4)LG3/2=lg3-lg2=b-a
1、对等式两边同除X,得X+X分之一=3 .两边平方得,x²+x²分之一+2=92、 原式=2X(x²-3X)+3x²-7X+2009=2X*(-1)+3(x²-3X)+2X+2009= - 3+ 再问： 第二题第一步化简的依据是什么 再答： 要凑出已
（1）原式=m(m^2+2m)+2008 =m(1+m)+2008 =m^2+m+第二题我没算出来,题没写错吧? 再问： 我就是要第二题 再答： 好吧，我不会。再问： 那我给你五分吧
sinx+cosx=√2(sinx+cosx)^2=2sin2x=1cos2x=0sin^4x-cos^4x=(sin^2x+cos^2x)(sin^2x-cos^2x)=(sin^2x-cos^2x)=-cos2x原式＝-cos2x＝0
因为x+1/x=3所以（x+1/x）^2=9所以x^2+1/x^2=9-2=7所以x^2/x^4+x^2+1=1/x^2+x^2+1=8
（1）由（tanx)\(tanx-1)=-1得到tanx=1/2,所以sinx/cosx=1/2,即cosx=2sinx,代入(sinx-3cosx)\(sinx+cosx),得到结果是-5/3(2)把cosx=2sinx代入sin^2x+cos^2x=1得到sinx=根号5/5,cosx=2倍根号5/5,所以sin^
X^2-3X+1=0∴x-3+1/x=0 两边除以x∴x+1/x=3 两边平方得x²+2+1/x²=9∴x²+1/x²=7 再两边平方得x^4+2+1/x^4=49∴x^4+1/x^4=47 x²+1/x²=7∴x²-2+1/x²=7-2∴
m-3>=06-2m>=0∴m=3把m=3代入得n=12∴√mn=√36=6√m×√n=√mn=6
（1）x²+1/x²=x²+2+1/x²-2=(x+1/x)²-2=3²-2=7,（2）x³+1/x³=(x+1/x)(x²-1+1/x²)=3×(7-1)=18,（3）x²/x的四次方+x²+1=1
解答绝对值≥0∴a-2=0a=2b-3=0b=3c=1∴（1）a-b-c=2-3-1=-2(2)|a+b|-|c|=|5|-1=4
X^1/2+X^-1/2=√￣x+1/√￣x=√￣[（√￣x+1/√￣x）^2]=√￣（x+1/x+2)=√￣(3+2)=√￣5X^2+X^-2 =(x+1/x)^2-2=3^2-2=7因为(x-1/x)^2=x^2+1/x^2-2=(x+1/x)^2-4=3^2-4=5所以x-1/x=√￣5所以X^2-X^-2 =(