x y=36;25*2x=40y 二元一次方程解-搜答案-智慧作业帮

(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=40+20x+y-2xy=60(x-xy)+(xy-y)=40+(-20)x-xy+xy-y=40-20x-y=20

x-xy=8(1)xy-y=-9(2)则有(1)-(2):X-XY-XY+Y=X+Y-2XY=8-(-9)=17

[（xy+2）（xy-2）-2x2y2+4]÷（xy）=（x2y2-4-2x2y2+4）÷xy=-x2y2÷xy=-xy,当x=10,y=1/25时,原式=-2/5．

[(xy+2)(xy-2)-2x^2y^2+4]/(xy)=[(x^2y^2-4)-2x^2y^2+4]/(xy)=-x^2y^2/xy=-xy=-10*(-1/25)=2/5

x^2+xy+y=14y^2+xy+x=28两式相加x^2+y^2+2xy+x+y=42(x+y)^2+(x+y)-42=0(x+y-6)(x+y+7)=0x+y=6或x+y=-7

(-3x^y+2xy)-(4x^+xy)=-3x^y+2xy-4x^-xy=-3x^y+xy-4x^所以填上-3x^y+xy-4x^

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60希望满意!

(x-xy)+(xy-y)=40-20x-xy+xy-y=20x-y=20(x-xy)-(xy-y)=40-(-20)x-xy-xy+y=60x+y-2xy=60

（1）∵xy+x=-1①，xy-y=-2②，∴①-②得x+y=1；（2）先把xy+x=-1，xy-y=-2的值代入代数式，得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8

x^2-xy=60,xy-y^2=40,x^2-y^2=(x^2-xy)+(xy-y^2)=60+40=100x^2-2xy+y^2=(x^2-xy)-(xy-y^2)=60-40=20

7或者-8再问：求过程^_^再答：两个等式两边相加

x^2-xy=60为（1）式xy-y^2=40为（2）式则x^2-y^2=（1）+（2）=60+40+100x^2.-2xy+y^2=（1）-（2）=60-40=20再问：为什么再答：==！这个应该显

2(x+xy)-[(xy-3y)-x]-(-xy)=2x+2xy-xy+3y+x+xy=3x+3y+2xy=3（x+y）+2xy=3*（-2）+2*3=0

∵x²-xy=60;xy-y²=40.x(x-y)=60;y(x-y)=40x-y=60/x;y*60/x=40y/x=2/3y=2/3x;x=3/2yx²-xy=60x

∵x-y=4xy，∴2x+3xy-2yx-2xy-y=2(x-y)+3xyx-y-2xy=8xy+3xy4xy-2xy=112．故答案为：112．

X²+2xy+y²/xy乘x²-2xy+y²/xy+y²=(x+y)²/xy×(x-y)²/y(x+y)=(x+y)(x-y)&#

x^2-xy与xy-y^2相比各个式子提取公因式（x-y）求得x/y=3/2两个等式相加就是x^2-y^2=100因式分解（x+y）*（x-y）=100将x=3y/2带入求得y的平方是20可以x的平方

[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+

x-y=(x-xy)+(xy-y)=40+(-20)=20x+y-2xy=(x-xy)-(xy-y)=60