初二数学:已知x^2-xy-2y^2=0,且x≠0,y≠0,求代数式x^2-2xy-5y^2\/x^...
=(-xy-3y^2)\/(3xy+7y^2)因为y≠0 原式=(-x-3y)\/(3x+7y)当x=2y时,原式=(-2y-3y)\/(6y+7y)= -5\/13 当x= - y时,原式=(y-3y)\/(-3y+7y)= -1\/2
...2y平方等于零,且x不等于零,y不等于零,求代数式x平方加二xy+5 y...
xy≠0。(x一y/2)^2一(3y/2)^2=0,(x+y)(x一2y)=0。①当x=一y时,x^2+2xy+5y^2 =y^2一2y^2+5y^2=4y^2;②当x=2y时,x^2+2xy+5y^2=4y^2+4y^2+5y^2 =13y^2。
已知x⊃2;-xy-2y⊃2;=0,且x、y不等于0,求代数式x⊃2;-2xy-5y...
解:∵x²-xy-2y²=0 ∴(x+y)(x-2y)=0 ∴x=-y或x=2y ∵且x、y不等于0 ∴1、把x=-y代入式子可得:(x²+2xy-5y²)\/(x²-2xy-5y²)=(y²-2y²-5y²)\/(y²+2y²-5y²)=3 2、把x=2y代入式子可得:(...
x^2y^2-xy-2y^2=0,求x\/y
=x^2-2xy+xy-2y^2=0 =x(x+y)-2y(x+y)=0 =(x-2y)(x+y)=0 所以记得x=2y 或者 x=-y 所以x\/y=2 或者 x\/y=-1 望采纳!
已知xy>0,且x^2-8y^2=2xy,求5x-2y\/x+2y
x^2-xy-2y^2=0 (x-2y)(x+y)=0 x=2y或x=-y (x^2-2xy-5y^2)\/(x^2+2xy+5y^2)=(x^2-xy-2y^2-xy-3y^2)\/(x^2-xy-2y^2+3xy+7y^2)=(-xy-3y^2)\/(3xy+7y^2)因为y≠0 原式=(-x-3y)\/(3x+7y)当x=2y时,原式=(-2y-3y)\/(6y+7y)= -5\/13 当x= - y时...
已知x,y为实数,且x^2+XY+Y^2-2=0,求x^2-xy+y^2的最大值和最小值
设x^2-xy+y^2=k,联立x^2+xy+y^2=2解得:x^2+y^2=(k+2)\/2,2xy=2-k 又x^2+y^2±2xy=(x±y)^2≥0 所以(k+2)\/2±(2-k)≥0 解得2\/3≤k≤6 所以x^2-xy+y^2最大值为6,最小值为2\/3
...Y,且X^2-X=5 , Y^2-Y=5,求代数式X^2-XY+Y^2的值
解:∵X≠Y,且X²-X=5 , Y²-Y=5 ∴可将X、Y看成方程m²-m=5的两个根 ∵m²-m-5=0 ∴X+Y=1 XY=-5 ∴X²-XY+Y²=(X+Y)²-3XY =1²-3×(-5)=1+15 =16
求证:无论x、y取何值,代数式x^2-xy+y^2-2x+5\/2的值总为非负数
配方 x²-xy+y²-2x+5\/2 =(1\/2)(2x²-2xy+2y²-4x+5)=(1\/2)[(x²-2xy+y²)+(x²-4x+4)+y²+1]=(1\/2)[(x-y)²+(x-2)²+y²+1]∵(x-y)²≥0,(x-2)²≥0,y²≥0 ∴(x-y)&...
已知x≠y,x^2-x=3,y^2-y=3,求代数式x^2+xy+y^2的值
x^2-x=3,y^2-y=3 可知:x^2-x-3=0 y^2-y-3=0 所以x、y是方程a^2-a-3=0的两个不等的根。有x*y=-3,x+y=1 则 (x+y)^2=1=x^2+y^2+2xy 结合x*y=-3可以知道 x^2+xy+y^2=x^2+y^2+2xy-x*y=1-(-3)=4 ...
已知x2-xy=5,xy-y2=-3,求代数式x2-y2和x2-2xy+y2的值。这样。
x2-y2 =x2-xy+xy-y2 =5-3 =2 x2-2xy+y2 =x2-xy-(xy-y2)=5-(-3)=8