x²-[5x²-4y] -3[ y-x²]
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原式=(15x²+16xy+4y²+x²-4y²)/4x=(16x²+16xy)/4x=4x+4y此题是利用多项式的乘法及平方差公式进行计算的,希望你
|2-x|+|x-y+4|=02-x=0,x-y+4=0x=2,2-y+4=0,y=63(x-y)-5(x-y)^2-3(x-y)+(y-x)^2+4(x+y)^2+3(y-x)=-4(x-y)^2+
原式=(9x²+24xy+16y²-4x²+y²-5x²+6xy-y²)÷(-2y)=(30xy+16y²)÷(-2y)=-15x
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=4x-2y-z-5x[6y-(x+y)]+x-(3y-10z)=4x-2y-z-30xy+5x²+5xy+x-3
-x^y^7+y^8)(这里有问题,x的几次方?再问:就是一次方再答:(x^8-x^7y+x^5y^3-x^4y^4+x^3y^5-xy^7+y^8)(x^2+xy+y^2)=(x^8-x^7y+x^
(3x-y)²-2x(4x-5y)-(x+2y)²=9x²-6xy+y²-8x²+10xy-x²-4xy-4y²=-3y²
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+
(x+2)(x+3)=x²+5x+6(x-4)(x+1)=x²-3x-4(y+4)(y-2)=y²+2y-8(y-5)(y-3)=y²-8y+15x²
楼上的全错,{(3x+y)(3x-y)-(x-5y)(5x-y)-(x-2y)²}÷(-4x)={9X²-(5x²-xy-25x²+5y²)-(x&s
=25x²-9y²+9x²-24xy+16y²=34x²-24xy+7y²
(y-x)(x-y)^3(y-x)^5=(y-x)[-(y-x)^3](y-x)^5=-(y-x)^(1+3+5)=-(y-x)^9=(x-y)^9
解题思路:对于这种等式一定可以化成平方相加的形式,这里面要使用到完全平方公式。解题过程:
(x-y)^5*(y-x)^3*(x-y)^2=-(x-y)^5*(x-y)^3*(x-y)^2=-(x-y)^(5+3+2)=-(x-y)^10
(5x+3y)(3y-5x)-(4x-y)(4y+x)=(3y)^2-(5x)^2-(4x^2+15xy-4y^2)=9y^2-25x^2-4x^2-15xy+4y^2=13y^2-15xy-29x^
(3x-4y)(5x-4y)-(7x-8y)(6y-5x)=15x²-12xy-20xy+16y²-(42xy-35x²-48y²+40xy)=15x²
[(3x+2y)(3x-2y)-(x+2y)(5x-2y)]/4x=[(9x^2-4y^2)-(5x^2+8xy-4y^2)]/4x=(9x^2-4y^2-5x^2-8xy+4y^2)/4x=(4x^
[(x+2y)²-(x+y)(3x-y)-5y²]÷(2x),其中x,y满足x²+4x+y²-y+17/4=0将x²+4x+y²-y+17/
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