求3(1+x^2)y'+2xy=2xy^4妻子满足我的绿帽需求条件y(0)=1/2的求特解

(1)已知x^2-3x-1=0 求x^2+1/x^2 (2)若x+1/x+2÷x+3/x+4有意义 求x范围 (3)已知2/x-2/y=6求 3x-2xy-3y/2y+xy-2x (4)已知2a-8/a+3的值时整数,求a的整数值!_百度作业帮
(1)已知x^2-3x-1=0 求x^2+1/x^2 (2)若x+1/x+2÷x+3/x+4有意义 求x范围 (3)已知2/x-2/y=6求 3x-2xy-3y/2y+xy-2x (4)已知2a-8/a+3的值时整数,求a的整数值!
(1)已知x^2-3x-1=0 求x^2+1/x^2 (2)若x+1/x+2÷x+3/x+4有意义 求x范围 (3)已知2/x-2/y=6求 3x-2xy-3y/2y+xy-2x (4)已知2a-8/a+3的值时整数,求a的整数值!
(1)已知x^2-3x-1=0 求x^2+1/x^2x^2-3x-1=0x-3-1/x=0x-1/x=3(x-1/x)^2=9x^2+1/x^2-2=9x^2+1/x^2=11(2)若x+1/x+2÷x+3/x+4有意义 求x范围x+1/x+2÷x+3/x+4=[(x+1)*(x+4)]/[(x+2)*(x+3)]上式有意义,分母不等于0x+2并不等于0,x不等于-2,同时x+3并不等于0,x不等于-3.(3)已知2/x-2/y=6,求 3x-2xy-3y/2y+xy-2x 2/x-2/y=61/X-1/Y=3(y-x)/xy=3y-x=3xy(3x-2xy-3y)/(2y+xy-2x)=[-3(y-x)-2xy]/[2(y-x)+xy]=(-3*3xy-2 xy)/(2*3xy+xy)=-11xy/7xy=-11/7 (4)已知2a-8/a+3的值时整数,求a的整数值!设(2a-8)/(a+3)=k,k为整数2a-8=ka+3k-a(k-2)=8+3ka=-(8+3k)/(k-2)a=-[3(k-2)+14]/(k-2)a=-3-14/(k-2),若a为整数,则-14/(k-2)为整数,14能被(k-2)整除取相应k值,代入:k=0,a=-3-14/(-2)=4k=1,a=-3-14/(-1)=11k=3,a=-3-14/1=-17k=4,a=-3-14/2=-10k=9,a=-3-14/7=-5k=16,a=-3-14/14=-4k=-5,a=-3-14/(-7)=-1k=-12,a=-3-14/(-14)=-2
(1)已知x^2-3x-1=0 求x^2+1/x^2x^2-3x-1=0x-3-1/x=0x-1/x=3(x-1/x)^2=9x^2+1/x^2-2=9x^2+1/x^2=11 (2)若x+1/x+2÷x+3/x+4有意义 求x范围 x+1/x+2÷x+3/x+4=[(x+1)*(x+4)]/[(x+2)*(...求曲线积分∫L(x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy,其中L是沿着椭圆x^2/4+y^2/4=1从A(2,0)B(-2,0)的一段弧结果是等于-(16/3)吗_百度作业帮
求曲线积分∫L(x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy,其中L是沿着椭圆x^2/4+y^2/4=1从A(2,0)B(-2,0)的一段弧结果是等于-(16/3)吗
求曲线积分∫L(x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy,其中L是沿着椭圆x^2/4+y^2/4=1从A(2,0)B(-2,0)的一段弧结果是等于-(16/3)吗
可以求得原函数U(x,y)=x^3/3+x^2*y-x*y^2-y^3/3+C.分别代入(2,0)跟(-2,0),作差得到结果为-(16/3),如楼主所言.(1)已知m-n=3,mn=1,求m^2+n^2的值(2)(-1/2)^2-2/3×9/8+2^0+(-1)^3+(4/3)^-1(3)(1/4a^2b)(-2ab^2)^2÷(-0.5a^4b^5)(4)化简,求值(x+y)(x-y)+(x-y)^2-(6x^2y-2xy^2)÷2y其中x=2,y=1_百度作业帮
(1)已知m-n=3,mn=1,求m^2+n^2的值(2)(-1/2)^2-2/3×9/8+2^0+(-1)^3+(4/3)^-1(3)(1/4a^2b)(-2ab^2)^2÷(-0.5a^4b^5)(4)化简,求值(x+y)(x-y)+(x-y)^2-(6x^2y-2xy^2)÷2y其中x=2,y=1
(1)已知m-n=3,mn=1,求m^2+n^2的值(2)(-1/2)^2-2/3×9/8+2^0+(-1)^3+(4/3)^-1(3)(1/4a^2b)(-2ab^2)^2÷(-0.5a^4b^5)(4)化简,求值(x+y)(x-y)+(x-y)^2-(6x^2y-2xy^2)÷2y其中x=2,y=1
(1)mn=1m-n=3两边平方m^2-2mn+n^2=9所以m^2+n^2=9+2mn=9+2*1=11 (2)(-1/2)^2-2/3×9/8+2^0+(-1)^3+(4/3)^-1 =1/4-3/4+1+(-1)+3/4=1/4(3)(1/4a^2b)(-2ab^2)^2÷(-0.5a^4b^5) =(-1/2a^3b^3)÷(-0.5a^4b^5)=1/(ab^2)(4)(x+y)(x-y)+(x-y)^2-(6x^2y-2xy^2)÷2y=x^2-y^2+x^2-2xy+y^2-3x^2+xy=-x^2-xy=-2^2-2*1=-6
(m-n)^2=9m^2+n^2-2mn=9m^2+n^2=9+2mn=9+2=11已知4(x+0.5)^2+│y+1│=0,求4x^2y-2/3xy^2+4x^2y+2-8x^2y+1/2xy^2的值_百度作业帮
已知4(x+0.5)^2+│y+1│=0,求4x^2y-2/3xy^2+4x^2y+2-8x^2y+1/2xy^2的值
已知4(x+0.5)^2+│y+1│=0,求4x^2y-2/3xy^2+4x^2y+2-8x^2y+1/2xy^2的值
4(x+0.5)^2+│y+1│=0,则x+0.5=0,y+1=0,x=-0.5,y=-1 4x^2y-2/3xy^2+4x^2y+2-8x^2y+1/2xy^2=-2/3xy^2+2+1/2xy^2=-1/6xy^2+2=1/6*0.5+2=2又1/12=2xy+2,此时x=y.将x=y代回圆方程,解出x=(7+根号7">
平面上有两点A(-1,0),B(1,0),点P在圆周(x-3)^2+(y-4)^2=4上,求使AP^2+BP^2最小值时点P的坐标已经知道作法和正确答案但是想问一下,问什么AP^2+BP^2=2(x^2+y^2)+2>=2xy+2,此时x=y.将x=y代回圆方程,解出x=(7+根号7_百度作业帮
平面上有两点A(-1,0),B(1,0),点P在圆周(x-3)^2+(y-4)^2=4上,求使AP^2+BP^2最小值时点P的坐标已经知道作法和正确答案但是想问一下,问什么AP^2+BP^2=2(x^2+y^2)+2>=2xy+2,此时x=y.将x=y代回圆方程,解出x=(7+根号7
平面上有两点A(-1,0),B(1,0),点P在圆周(x-3)^2+(y-4)^2=4上,求使AP^2+BP^2最小值时点P的坐标已经知道作法和正确答案但是想问一下,问什么AP^2+BP^2=2(x^2+y^2)+2>=2xy+2,此时x=y.将x=y代回圆方程,解出x=(7+根号7)/2和(7-根号7)/2.为什么这样做和答案不一样, 错在哪里?
设P点坐标(x,y),P在圆周上,所以P满足(x-3)²+(y-4)²=4PA²=(x+1)²+y² PB²=(x-1)²+y²PA²+PB²=2x²+2y²+2把圆的方程展开x²-6x+9+y²-8y+16=4→ x²+y²+1=6x+8y-20PA²+PB²=4(3x+4y-10)∵3x+4y≥2√12xy=4√3xy且当3x=4y时∴代入圆的方程得P(21/5,28/5)
设p坐标:(3+2cosα,4+2sinα)(ap)^2+(bp)^2=(4+2cosα)^2+2(4+2sinα)^2+(2+2cosα)^2=60+24cosα+32sinα=60+8/5×(3/5cosa+4/5sina)=60+8/5sin(α+β),其中β=arctan3/4当sin(α+β)=-1时取到最小值,即sina=-4/5,cos=-3/5所以p坐标:(9/5,12/5...
均值不等式用之后必须得到定值,你得到2xy+2不是定值,所以得不到正确答案。

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