y=1 (2x-y²)的解-搜答案-搜搜做题作业网

1/40再问：过程再答：我想让您再发一遍原式行吗？((x+y)的平方-(x-y)的平方-2y(x-1/2y))/4y里的“x-1/2y”是(x-1)/2y还是x-(1/2y)?再问：第二个再答：(x+

x+y/x-y=1/2取倒数x-y/x+y=2所以x-y/x+y-2x+2y/x-y=x-y/x+y-2(x+y/x-y)=2-2×1/2=2-1=1

[2x的平方-(x+y)(x-y)][(-x-y)(y-x)+2y的平方]=[2x²-(x²-y²)][(x+y)(x-y)+2y²]=(2x²-x&

y'=p,y''=dp/dx,xp'+xp^2-p=0p'=(-xp^2+p)/x,p'-1/x*p=-p^2伯努利方程,换元求出p,再求y

(2x-y)(2x+y)+(2x-y)(y-4x)+2y(y-3x)=4x^2-y^2+2xy-8x^2-y^2+4xy+2y^2-6xy=-4x^2=-4(-1/4)^2=-1/4

x+y/3+x-y/2=63(x+y)-2(x-y)=282x+2y+3x-3y=363x+3y-2x+2y=285x-y=36x+5y=285x-y=365x+25y=14026y=104y=4x=

56

(1)x-2y=1,2x-4y=3(2)x-2y=1,2x-y=3(3)x-2y=1,2x-4y=2方程组（1）左式比例相同1:2,且与左式的比例1:3不等,是两个矛盾的方程,故无解方程组（3）左式比

第二个方程化为Y=M+1-2X代入第一个并化为X=（M+3）/3>0得,M>-3,再把第一个化为X=M-1-2Y代入二得Y=（M-3）/3

原式=[x^2+4xy+4y^2-x^2+y^2-7y^2]/2y=[4xy-2y^2]/2y=2x-y代入x=-1,y=-2原式=-2-（-2）=0

x+y=1x-y=2(x+2y)(x-2y)-(2x-y)(-y-2x)=(x+2y)(x-2y)+(2x-y)(y+2x)=x²-4y²+4x²-y²=5x&

我用x2表示的x的平方1,(x-y)2=x2-2xy+y2,(x+y)(x-y)=x2-y2,所以原式=(2x2-2xy)/2x=x-y=12,=xy+y2+x2-y2-x2=xy=-13,=x2-x

(x-y)[(x+y)+x-y]+4y-2y^2=2x(x-y)+4y-2y^2=2[x^2-xy+2y-y^2]所以=2[1-2+4-4]/8=-1/4

解(x-y)(x+y)-(x-2y)²+x(3x-5y)-(x-y)(x-2y)=(x²-y²)-(x²-4xy+4y²)+(3x²-5xy

即xy'-y=x^3即(xy'-x'y)/x^2=x即(y/x)'=xy/x=1/2x^2+cy=x(1/2x^2+c)；c为常数

7y(x-3y)²-2(3y-x)³=7y(x-3y)²+2(x-3y)³=(x-3y)²[7y+2(x-3y)=(x-3y)²(7y+2X

这是线性规划的题目根据x-y≥-1,x+y≥1,2x-y≤1画图可以发现满足的点位于三条直线所围成的三角形内.x-y=-1与2x-y=1交点为x=2,y=3即交点为(2,3)Z=(x-2y)/(x+y

(x-2y)/(x+2y)=3取倒数(x+2y)/(x-2y)=1/3所以原式=(1/3)[(x-2y)/(x+2y)]-3[(x+2y)/(x-2y)]=(1/3)×3-3×(1/3)=0

【（x+2y）2-(x+y)(3x-y)-5y2】÷（2x）=[x^2+4xy+y^2-(3x^2+2xy-y^2)-5y^2]/(2x)=[-2x^2+2xy]/(2x)=y-x=1/2-2=-3/