已知x 2y=0,求分式x的平方-xy分之2xy y的平方的值
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x2y+xy2-x-y=xy(x+y)-(x+y)=(x+y)(xy-1)∵x+y=-5,xy=7,∴原式=-5×(7-1)=-30.
原式=(x^2-2x+1)+(y^2-4y+4)=(x-1)^2+(y-2)^2=0,所以x=1,y=2,代入问题得结果为3/2
分式2x-m/x+n,当x=5时,分式无意义5+n=0n=-5当x=-1时,分式的值为02*(-1)-m=0m=-2会了吧?
x-3y=0且xy≠0,所以x=3y(x≠0,y≠0)带入分式,(9y-3y+y)/(9y+3y+y)=7/13
1)由X^2-3X+1=0可知x不等于0两边除以x得X+1/X=3X^2/(X^4+x^2+1)=1/(x^2+1+1/X^2)=1/[(X+1/X)^2-2+1]=1/(9-1)=1/82)由X^2
x^2+y^2-4x-6y+13=0(x^2-4x+4)+(x^2-6y+9)=0(x-2)^2+(y-3)^2=0所以(x-2)^2=0,(y-3)^2=0x-2=0,y-3=0x=2,y=3xy/
由已知1,分母为0,b=6,由已知2,a=-5.则6a+9b=6*6+9*(-5)=-9
x+1/x=32边平方x的平方+x的平方分之一+2=9x的平方+x的平方分之一=7
x3+y3-x2y-xy2=(x+y)(x2-xy+y2)-xy(x+y)=(x+y)(x2-2xy+y2)=(x+y)(x2+2xy+y2-4xy)=(x+y)[(x+y)2-4xy]=10×(10
x的绝对值-1分之x的平方-7x-8=0(x+1)(x-8)/(lxl-1)=0x=-1x=8由于x≠±1所以x=8
∵3X+4Y=0(X不等于0)∵y=-3x/4∴原式=(-3x²/2+9x²/16)/(x²+3x²/4)=-15/16*4/7=-105/4再问:这分式怎么X
(x+y)(x-y)-y^2+(x-y)^2-(6x^2y-2xy^2)/(2y)=X^2-y^2-y^2+X^2+y^2-2xy-3x^2+xy=-x^2-y^2-xy=-(x^2+y^2+xy-3
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
4+3/x=0x=-3/4(2x-x^2)/(x^3+x^2)*(x^2+2x+1)/(x^2-4x+4)=[x(2-x)/x²(x+1)]*[(x+1)²/(x-2)²
x的平方+2xy+y的平方分之x的平方-y的平方=(x+y)(x-y)/(x+y)²=(x-y)/(x+y)∵x+y=2,x-y=1∴原式=1/2
原式=5xy2-2x2y+3xy2-2x2y=8xy2-4x2y,∵(x-2)2+|y+1|=0,∴x-2=0,y+1=0,即x=2,y=-1,则原式=16+16=32.
(2x^2+3xy-2y^2)/(x^2+3xy+2y^2)={(2x-y)(x+2y)}/{(x+y)(x+2y)}=(2x-y)/(x+y)=02x-y=0
∵x2-y2=xy,∴原式=x2y2+y2x2=x4+y4x2y2=(x2−y2)2+2x2y2x2y2=3x2y2x2y2=3.再问:先化简2a+1/a²-1÷a²-a/a
x²-x=7y²-y=7相减x²-x-y²+y=0(x+y)(x-y)=x-yx-y≠0约分x+y=1x²-x=7y²-y=7相加x&sup