x² x-4y² 2y=

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x² x-4y² 2y=
化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1

原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x

[(x+2y)(x-2y)-(x-2y)^2+8y(x+y)]除以4x=

[(x+2y)(x-2y)-(x-2y)^2+8y(x+y)]/4x=[x^2-4y^2-x^2-4y^2+4xy+8xy+8y^2]/4x=12xy/4x=3y

2x-y=2,求[(x²+y²)-(x-y)²+2y(x-y)]÷4y

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

x+y/2+x-y/3=6,4(x+y)-3(x-y)=-20

由(1)得3x+3y+2x-2y=365x+y=36(3)由(2)得4x+4y-3x+3y=-20x+7y=-20(4)(3)×7-(4)得34x=272∴x=8把x=8代入(3)得y=-4∴x=8y

{(x+y)/2+(x-y)/3=6 4(x+y)-3(x-y)=-20

{(x+y)/2+(x-y)/3=63(x+y)+2(x-y)=36(1)4(x+y)-3(x-y)=-20(2)由(1)*3+(2)*2得9(x+y)+6(x-y)+8(x+y)-6(x-y)=36

已知2x-y=10,求[(x²+y²)-(x-y)²+2y(x-y)]/4y

先化简[(x²+y²)-(x-y)²+2y(x-y)]/4y[(x²+y²)-(x-y)²+2y(x-y)]/4y=[x²+y&s

3(x+y)-4(x-y)=11,2(x-y)+5(x+y)=27

化简得:-x+7y=11①7x+3y=27②①式×7得:-7x+49y=77③②+③得:52y=104∴y=2代入①得:x=3∴x=3,y=2再问:亲,是代入法哦!再答:代入法①式得3x+3y-4x+

已知x/y=2,求2x(x+y)-y(x+y)/4x²-4xy+y²

原式=(2x-y)(x+y)/(2x-y)^2=(x+y)/(2x-y)x/y=2x=2y原式=3y/3y=1

x^2-y^2/x+y-4x(x-y)+y^2/2x-y,

(x^2-y^2)/(x+y)-(4x(x-y)+y^2)/(2x-y)=(x-y)(x+y)/(x+y)-(4x^2-4xy+y^2)/(2x-y)=(x-y)-(2x-y)^2/(2x-y)=(x

数学题……555{(3x+y)(3x-y)-(x-5y)(5x-y)-(x-2y)²÷(-4x),x=-2.y

楼上的全错,{(3x+y)(3x-y)-(x-5y)(5x-y)-(x-2y)²}÷(-4x)={9X²-(5x²-xy-25x²+5y²)-(x&s

若(x*x+y*y)(x*x+y*y)-4x*x*y*y=0,求代数式(x*x+5xy+y*y)/(x*x+2xy+y*

(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=

4(x+y)-3(x-y)=-20,2/x+y+3/x-y=6

第二个方程是不是写错了2/(x+y)+3/(x-y)=6是这样吗

已知x*x+4x+y*y-2y+5=0,则x*x+y*y=?

X^2表示平方X^2+4X+4+Y^2-2Y+1=0(X+2)^2+(Y-1)^2=0因为平方大于=0所以X+2=0Y-1=0X=-2Y=1X^2+Y^2=5

已知x*x-4xy+4y*y=0 求[2x(x+y)-y(x+y)]/(4x*x-4xy+y*y)的值?

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

{3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1

3(x+y)-4(x-y)=4(x+y)/2+(x-y)/6=1令a=x+y,b=x-y3a-4b=4(1)a/2+b/6=1则3a+b=6(2)(2)-(1)5b=2b=2/5a=(6-b)/3=2

已知4x=9y求(1)x+y/y (2)y-x/2x

4x=9yx=9/4*y(1)(x+y)/y=[(9/4)y+y]/y=(9/4+1)y/y=9/4+1=13/4(2)(y-x)/2x=[y-(9/4)y]/[2*(9/4)y]=(1-9/4)y/

y*y*y*y+2x*x*x*x=4x*x*y 求x等于几,y等于几,写出所有选项!

把原方程整理得:2x^4-4x^2y+y^4+1=0则这个关于x^2的一元二次方程有实数解,故得:(-4y)^2-8(y^4+1)≥0即有:(y^2-1)^2≤0当且仅当y^2-1=0时,上述方程有实

已知x²+y²+5=2x+4y,求【2x²-(x-y)(x-y)】【(x+y-1)(x-y

1,-3再问:过程。。。再答:★(x²-2x)+(y²-4y)=5★(x-1)²+(y-2)²=1+4-5★(x-l)²=0,(y-2)²=

已知x=1/3,y=-1/2,求代数式x-(x+y)+(x+2y)-(x+3y)+(x+4y)-(x+5y)+...-(

原式=x-x+x-x+……-x+(2-1+4-3+5-4+……+2008-2007-2009)y=0+(1×1004-2009)y=-1005y=1005/2

(x+y)(x+2y)(x+3y)(x+4y)=-40

我把方法告诉你,最后的答案你自己做吧,很容易.(x+y)(x+2y)(x+3y)(x+4y)=-40(x+y)(x+4y)(x+2y)(x+3y)=-40(x^2+5yx+44)(x^2+5yx+66