帮忙matlab求解一下这个符号方程 ,要求出y=.y=x+(480m+15n+20x)*(k-x)/k+(k-x-(480m+15n+20x)*(k-x)/k)*(1-(k-x-(480m+15n+20x)*(k-x)/k-a*(k-x-(480m+15n+20x)*(k-x)/k)/k)/(k-x-(480m+15n+20x)*(k-x)/k))^((480m+15n+20x)*(k-x)/k)y和x是变量，其他字母是常量
分类：数学
必须知道 a,m,n,x,k的值或者取值范围,才可以求出y啊.如果你知道这五个变量的情况,在matlab里直接敲入这个方程就可以了
3乘以－的根号2平方,再减去5乘以－的根号二－3（√22）＋5（√2）＋2，我把题目的正负之类的化简了一下
3×（-√2）?－5×（-√2）=3×2+5√2=6+5√2
已知向量a=(2cosX/2,tan(X/2+π/4)),b=(根号二sin(X/2+π/4)),tan=(X/2—π/4)令f(x)=a*b,求函数f（x）的最大值、最小正周期、并写出f（x）在【0,π】的单调区间
a=(2cosX/2,tan(X/2+π/4))=（2cosx/2,（1+tanx/2）/（1-tanx/2））b=(根号二sin(X/2+π/4)),tan=(X/2—π/4)=（sinx/2+cosx*2,（tanx/2-1）/（1+tanx/2））f(x)=a*b=sinx+cosx+1-1=√2sin（x+π/4）所以f（x）最大值是√2,最小正周期是T=2πx属于[0,π] 得到x+π/4属于[π/4,5π/4]得到f（x）在[π/4,π/2）上递增,在[π/2,5π/4]上递减
matlab中,有一个三维图像,如何沿着两个坐标轴得到剖面图?有什么函数?最好能写个完整的表达式,用法详细点,我是matlab菜鸟先谢过,这个方法很好,但是不知道有没有写代码的方法,因为这是作业,要交给老师看的.
算筹公元429 年,祖冲之诞生在范阳郡遒县（今河北省涞源县）的一个士大夫家庭．他的祖父、父亲都很喜欢数学．受家庭环境的影响,祖冲之从儿时起,就对数学着迷．每当父辈们用”算筹”来计算时,他就瞪着好奇的大眼睛,默默地瞅着那些”算筹”．渐渐地,他也能得心应手地摆弄这些用来计算的小竹棍了．随着年龄的增长,祖冲之已不满足於那些简单的运算,他开始研究前人的成果,希望在此基础上有更大的突破．一天,祖冲之得到了一本刘徽作注的《九章算术》．他如获至宝．上朝归来,便躲在书斋里潜心阅读．随后不久,祖冲之便开始了他的计算工作．当时,没有计算机等先进的计算工具,所有的只是一些作为算筹的小竹棍．祖冲之便利用这原始的计算工具,每天在公务之余不停地计算着．从12 边形、24 边形、48 边形、96 边形、192 边形、768 边形、1536 边形、到12288 边形,反复地运算．一根根小竹棍被摸得通红发亮,一双手被磨出了厚厚的老茧．经过多年不懈的努力,终於得出了比较精确的结论．3.1415926＜π＜3.1415927这个数值在当时的世界上是最精确的,直到一千年之后,才有人打破这个纪录．
因为 y=(2--m)x^(m^2--3)--4是一次函数,所以 2--m不等于0,且 m^2--3=1即：m不等于2,且 m=正负2,所以 m=--2.
如果函数f(x)的定义域为（0,正无穷大）,且f(x)为增函数,f(xy）=f(x)+f(y) (1)证明：f(x/y)=f(x)-f(y)(2)已知f（3）=1,且f(a)大于f(a-1）+2,求a的取值范围。
分析：（1）结合抽象表达式用x/y代替x,y不变,即可转化即可获得问题f(x/y）=f(x)-f(y)的解答；（2）首先利用数值的搭配计算f（9）=2,进而对不等式进行转化,然后结合函数y=f（x）是定义在（0,+∞）上的单调性,结合变形后的抽象函数即可获得变量a的要求,进而问题即可获得解答．（1）∵对一切x,y＞0满足f（x）+f（y）=f（xoy）,∴f(x/y)+f(y)=f(x/y×y)=f(x)因此,满足 f(x/y)=f(x)-f(y),（2）∵f（3）=1,∴2=f（3）+f（3）=f（9）；∵f（x）是定义在（0,+∞）上的增函数,∴f（a）＞f（a-1）+2,a-1＞0,a＞0,f[(a-1)o9]＜f(a)；a＞1,(a-1)o9＜a1＜a＜9/8,故a的取值范围（1,9/8）点评：本题考查的是抽象函数及其应用的综合类问题．在解答的过程当中充分体现了定义域优先的原则、特值的思想、转化的思想以及计算和解不等式组的能力．值得同学们体会和反思．.,.
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扫一下，分享更方便，购买更轻松Binomial theorem - Topics in precalculus
P R E C A L C U L U S
THE BINOMIAL THEOREM
THE BINOMIAL THEOREM shows how to calculate a power of a binomial -- (a + b)n -- without actually multiplying out.
For example, if we actually multiplied out the 4th power of (a + b) --
(a + b)4 = (a + b)(a + b)(a + b)(a + b)
-- then on collecting like terms we would find:
b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 .& .&
Notice: &The literal factors are all the combinations of a&and&b where the sum of the exponents is 4: &a4, &a3b, &a2b2, &ab3, &b4.
of each term is 4.
The first term is actually a4b0, which is a4& 1.
to &expand& (a + b)5, we would anticipate the following terms, in which the sum of all the exponents is 5:
(a + b)5 = &?&a5 + &?&a4b + &?&a3b2 + &?&a2b3 + &?&ab4 + &?&b5
The question is, What are the coefficients?
They are called the binomial coefficients. &In the expansion of (a + b)4, the binomial coefficients are
& Note the symmetry: &The coefficients from left to right are the same right to left.
The answer to the question, "What are the binomial coefficients?" is called the binomial theorem. &It shows how to calculate the coefficients in the expansion of (a + b)n.
The symbol for a binomial coefficient is . &The upper index n is the expon the lower index k indicates which term, starting with k = 0.
= 5, each term in the expansion of &(a + b)5 &will look like this:
k will successively take on the values 0 through 5.
(a + b)5 = a5 &+ &a4b &+ &a3b2 &+ &a2b3 &+ & ab4 &+ &b5
Note: &Each lower index
is the exponent of b. &The first term has k&=&0 because in the first term, b appears as b0, which is 1.
Now, what are these binomial coefficients,
The theorem states that the binomial coefficients are none other than the , nCk .
& (a + b)5
5C0a5 + 5C1a4b + 5C2a3b2 +
5C3a2b3 + 5C4ab4 + 5C5b5
1a5 + a4b + a3b2 + a2b3 + ab4 + b5
a5 &+& 5a4b &+& 10a3b2 &+& 10a2b3 &+& 5ab4 &+& b5
The binomial coefficients here are
1& 5& 10& 10& 5& 1.
Note the symmetry.
The coefficient of the first term is always 1, and the coefficient of the second term is the same as the exponent of (a&+&b), which here is 5.
Note also that, when the
combinatorial number is written as factors, each coefficient contains the previous one. &The coefficient of a3b2 is the previous coefficient multiplied by 4 and divided by&2.
That is, it is multiplied by the exponent of the previous a, and divided by the present exponent of b.
The coefficient of a2b3 is the previous coefficient multiplied by 3 and divided by&3. &And so on. &We will see this in .
for the combinatorial numbers, here is the binomial theorem:
What follows the summation sign is the general term. &Each term in the sum will look like that -- the first term having k = 0; then k = 1, k&=&2, and so on, up to k = n.
Notice that the sum of the exponents &(n&&&k) + k, always equals n.
Example 1.
a) &The term &a8b4 &occurs in the expansion of what binomial?
Answer.&&&(a + b)12. &The sum of 8 + 4 is 12.
b) &In that expansion, what number is the coefficient of a8b4?
&Answer.&&&It is the combinatorial number,
12&&11&&10&&9&&&1&&2&&3&&4
Note again: The
lower index, in this case 4, is the .
This same number is also the coefficient of &a4b8, &since &12C8 = 12C4.
Example 2.&&& (a & b)5.
&Solution.&&&We found the
to be &1 &5& 10& 10& 5& 1. &The difference with (a & b) is that the signs of the terms will alternate:
(a & b)5 = a5 & 5a4b + 10a3b2 & 10a2b3 + 5ab4 & b5.
For, &a & b = a + (&b), &therefore each term will have the form
a5 & k(&b)k.
When k is even, (&b)k will be positive. &But when k is odd, (&b)k will be negative. &
Each odd power of b will have a minus sign.
Example 3.&&&In the expansion of &(x&&&y)15, calculate the coefficients of x3y12&and& x2y13.
&Solution.&&&The coefficient of& x3y12 &is positive
because the exponent of
y is even. &That coefficient is 15C12. &But 15C12 = 15C3, and so we have
15&&14&&13 &&&&1&&2&&3
The coefficient of &x2y13, on the other hand, is negative because the exponent of y is odd. &The coefficient is && 15C13
&=& &&15C2. &We have
15&&14&&1&&2
Example 4.&&& the first four terms of (a + b)n. &Do not use factorials.
&&&(a + b)n
+ &nC1an & 1b&
+ &nC2an & 2b2&
+ &nC3an & 3b3
+ an & 2b2
+ an & 3b3.
Notice: &Each coefficient is a factor of the next coefficient. &The coefficient n is a factor of
That in turn is a factor of
To construct the next coefficient, then, multiply the present coefficient by the exponent of a in that term --
-- namely n & 3:
n(n & 1)(n & 2)(n & 3)&&&&&&&&&1&&2&&3&&4
And divide it by 1 more than the exponent of b.
That is the coefficient of an & 4b4.
Example 5.&&& the binomial theorem to expand &(a + b)8.
&Solution.&&&The expansion will begin :
(a + b)8 = a8 + 8a7b
The first coefficient is always 1. &The second is always the exponent of the expansion, in this case 8.
The next coefficient can be constructed as described . &It will be the present coefficient, 8 --
(a + b)8 = a8 + 8a7b + 28a6b2
-- times the exponent of a in that term , 7, divided by 1 more then the exponent of b. &It will be 8&&7/2 = 28.
The next coefficient --
(a + b)8 = a8
-- is 28&&6, divided by 3:
28&&6 / 3 = 28&&2 = 56.
The next --
(a + b)8 = a8
-- is 56&&5, divided by 4:
56&&5 / 4 = 14&&5 = 70.
We have now come to the point of symmetry.& For, the coefficient of a3b5 is equal to the coefficient of a5b3, which is 56. &And so on for the remaining coefficients.
Here is the complete expansion:
a8&+&8a7b&+& 28a6b2&+&56a5b3&+&70a4b4&+& 56a3b5&+
28a2b6&+&8ab7&+&b8.
Example 6.&&&Write the 5th term in the expansion of (a + b)10.
&Solution.&&&In the 1st term, k = 0. &In the 2nd term, k = 1. &And so on.
The index k -- the exponent of b -- of each term is one less than the
of the term.
Thus in the 5th term, k = 4. &The exponent of b is 4. &The 5th term is
10&&9&&8&&71&&2&&3&&4
&a6b4&&=&&210 a6b4
This triangular array is called Pascal's Triangle. &Each row gives the combinatorial numbers, which are the binomial coefficients. &That is, the row &1&&2&&1& are the combinatorial numbers 2Ck, which are the coefficients of (a&+&b)2. &The next row,& 1 &3& 3& 1,& are the coefficients of (a + b)3; and so on. &
To construct the triangle, write 1, and below it write 1 &1. &Begin and end each successive row with 1. &To construct the intervening numbers, add the two numbers immediately above.
Thus to construct the third row, begin it with&1,
then add the two numbers immediately above: &1 + 1. &Write 2. &Finish the row with 1. &
To construct the next row, begin it with&1, and
the two numbers immediately above: 1 + 2. &Write 3. &Again, add the two numbers immediately above: &2 + 1 = 3. &Finish the row with 1.
Example 8.&&&Expand &(x & 1)6.
&Solution.&&&According to , the coefficients are
1& 6& 15& 20& 15& 6& 1.
In the binomial, x is &a&, and
&1 is &b&. &The
will alternate:
x6 & 6x5&&1
+ 15x4&&12
& 20x3&&13
+ 15x2&&14
x6 & 6x5 + 15x4 & 20x3 + 15x2 & 6x + 1.
Example 9.&&&Expand &(x + 2)3.
&Solution.&&&The coefficients are &. & x is &a&, and 2 is &b&.
x3 + 3x2&&2 + 3x&&22 + 23
x3 + 6x2 + 12x + 8
Problem 1.&&& the expansion of (a + b)n, each term has the form
where k successively takes on the value 0, 1, 2, . . ., n.
is the symbol for the binomial coefficient.
The binomial theorem is the statement that & = ?
To see the answer, pass your mouse over the colored area. To cover the answer again, click "Refresh" ("Reload").
The combinatorial number,&nCk.
Problem 2.&&&Use factorials to write the general term in the expansion of (a + b)n.
&&&&&&n!&&&&&&(n & k)! k!
Problem 3.&&&
a) & Calculate the coefficient of &a4b6 &in the expansion of (a + b)10.
10C6 = 10C4 = 210
b) & The coefficient of which other term is the same?&&
c) & In the expansion of &(a + b)n, the coefficient of &an & kbk& is the same
c) &&as the coefficient of which other term?&&
Problem 4.&&&Calculate the coefficient of
a) & x17y3 in the expansion of (x + y)20.&&
b) & x3y17 in the expansion of (x + y)20.&&
c) & x3y17 in the expansion of (x & y)20.&&
d) & x2y18 in the expansion of (x & y)20.&&
e) & x5y5 in the expansion of (x & y)10.&&
f) & x10 in the expansion of (x & 1)15.&&
Problem 5.&&& the first four terms of (x +
h)n. &Do not use factorials.
xn + nxn&1h +&
n(n & 1)&&1&&2
n(n & 1)(n & 2)&&&&&&1&&2&&3
Problem 6.&&&Compute the first four terms of each of the following.
a) & (a + b)15&&
15a14b + 105a13b2 + 455a12b3
b) & (x & 1)20&&
x20 & 20x19 + 190x18 & 1140x17
Problem 7.&&&Consider the expansion of (x&+&b)30.
a) & What is the exponent of b in the 1st term?&&
b) & What is the exponent of b in the 3rd term?&&
c) & In the 25th term?&&
d) & In the kth term?&&
e) & Write the fourth term, with its coefficient.&&
4,060x27b3
Problem 8.&&&Calculate each of the following.
a) & The third term of (a + b)11.&&
b) & The fifth term of (x & y)7.&&
c) & The tenth term of (x & 1)12.&&
& &d) & The fifteenth term of (1 +&
& &e) & The fourth term of (x &&
Problem 9.&&&Use Pascal's triangle to expand the following.
a) & (a + b)3&
= &a3 + 3a2b + 3ab2 + b3
= &a3 & 3a2b + 3ab2 & b3
= &x4 + 4x3y + 6x2y2 + 4xy3 + y4
= &x4 & 4x3y + 6x2y2 & 4xy3 + y4
= &x5 & 5x4 + 10x3 & 10x2 + 5x & 1
= &x5 + 10x4 + 40x3 + 80x2 + 80x + 32
= &8x3 & 12x2 + 6x & 1
= &1 & 7xy + 21x2y2 & 35x3y3 + 35x4y4 & 21x5y5 + 7x6y6 & x7y7
In the following Topic we will explain why the binomial coefficients
are the combinatorial numbers. That will constitute a proof of the binomial theorem.
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Lawrence Spector
Questions or comments?
Private tutoring available.已知函数f(x)=x^+1,g(x)为一次函数,且是增函数,若f [g(x)]=4x^-20x+26,求g(x)的解析式_百度知道
已知函数f(x)=x^+1,g(x)为一次函数,且是增函数,若f [g(x)]=4x^-20x+26,求g(x)的解析式
详细过程也要有。
b=5,k=2,b=-5.k=-2已知函数f(x)=x^+1,g(x)为一次函数,且是增函数,若f [g(x)]=4x^-20x+26,k=2,-5,所以y=2x-5,b^2+1=26b=5,-22kbx=-20x, kb=-10,求g(x)的解析式g(x)=kx+bf[g(x)]=[g(x)]^2+1=(kx+b)^2+1=k^2*x^2+2kbx+b^2+1=4x^-20x+26对照就可以解得k^2=4
采纳率：15%
你再把题目手抄一遍做个图片用百度hi发给我，我给你做
f(x)=x²+1由题意，设g(x)=kx+b，(k&0)，则f(g(x))=(kx+b)²+1=k²x²+2kbx+b²+1又f(g(x))=4x²-20x+26，比较系数，得k²=4，2kb= -20，b²+1=26，k&0∴k=2，b= -5，g(x)=2x-5.
不妨设g(x)=kx+b(k＞0)则f[g(x)]=f(kx+b)=(kx+b)²+1=k²x²+2kbx+b²+1，得到：k²=42kb=-20b²+1=26解得：k=2，b=-5∴g(x)=2x-5
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