4x^2 y^2 8x=12

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已知2x+y=4,求代数式[(x+y)2-(x-y)2-2y(x-12

原式=[x2+y2+2xy-x2-y2+2xy-2xy+y2]÷4y=(2xy+y2)÷4y=14(2x+y)=14×4=1.

[(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

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

{3X+2Y=22 5X+4Y=28 {4X+3Y=12 5X+2Y=11 {6X+5Y=21 5X+8Y=22 二元一

一、y=11-3x/2代入后一个方程得5x+4(11-3x/2)=28,解得x=16,y=11-3*16/2=-13二、y=4-4x/3代入后一个方程得5x+2(4-4x/3)=11,解得x=9/7,

已知2x-y/x+3y=2,求代数式4x-2y/x+3y减4x+12y除以2x-y

2x-y/x+3y=2得x+3y/2x-y=1/24x-2y/x+3y减4x+12y除以2x-y=2(2x-y)/x+3y-4(x+3y)/2x-y=2乘以2-4乖以1/2=4-2=2

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

已知2x-y/x+3y=2,求代数式4x-2y/x+3y减4x+12y/2x-y的值

答案是2.4x-2y/x+3y=(2x-y/x+3y)乘以2=4,4x+12y/2x-y=(x+3y/2x-y)乘以4=1/2x4=2,所以代数式等于4-2=2.

已知X.x+y.y-4x+2y+5=0,求4X.X-12X Y+9Y.y的值

∵x²+y²-4x+2y+5=0x²-4x+4+y²+2y+1=0(x-2)²+(y+1)²=0x-2=0y+1=0∴x=2y=-14x&#

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

解方程组:x+y/2+x-y/3=6 4(x+y)-5(x-y)=12

设a=x+y,b=x-y原方程组为:a/2+b/3=6,即3a+2b=36,即:12a+8b=1444a-5b=12,即:12a-15b=36相减得:23b=108,得:b=108/23,再得:a=1

若x-2y=5,则式子12-x+2y=.以知3x-4y=2x+y则x∶y=

12-x+2y=12-(x-2y)=12-5=73x-4y=2x+yx=5yx:y=5:1

{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/

x²+4y²-2x+12y+10=0求X,Y值

即(x²-2x+1)+(4y²+12y+9)=0(x-1)²+(2y+3)²=0所以x-1=0,2y+3=0所以x=1,y=-3/2

(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