(x+y)^2(x-y)^2-(x-y)(x+y)(x^2+y^2).
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(x+y)^2(x-y)^2-(x-y)(x+y)(x^2+y^2).
已知a,b,c分别为三角形ABC的三条边长,你能用因式分解的知识说明b^2+c^2-a^2+2bc一定是正数吗?
已知a,b,c分别为三角形ABC的三条边长,你能用因式分解的知识说明b^2+c^2-a^2+2bc一定是正数吗?
原式=[(x+y)(x-y)]^2-(x^2-y^2)(x^2+y^2)
=(x^2-y^2)^2-(x^4-y^4)
=x^4+y^4-x^4+y^4-2x^2y^2
=+2y^4-2x^2y^2
b^2+c^2-a^2+2bc
=(b+c)^2-a^2
=(b+c+a)(b+c-a)
三角形两边之和大于第三边
所以原式大于0
=(x^2-y^2)^2-(x^4-y^4)
=x^4+y^4-x^4+y^4-2x^2y^2
=+2y^4-2x^2y^2
b^2+c^2-a^2+2bc
=(b+c)^2-a^2
=(b+c+a)(b+c-a)
三角形两边之和大于第三边
所以原式大于0
1、x(x-y)(x+y)-x(x+y)^2
(1)(x^2/x)-y-x-y
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[(-x-y)(-x+y)-(x+y)^2-x(y-y^2)}÷1/2y
x^2-y^2/x+y-4x(x-y)+y^2/2x-y,
(x+y)^2(x-y)^2-(x-y)(x+y)(x^2+y^2)分解因式
化简:(x-3y)(x+y)-(x-2y)(x+2y)-(x-y)平方
【x-3y][x+y]-[x-2y][x +2y]-[x-y
[(3x+y)(x-y)-(2x+3y)(x-y)]*(x-y)
有这样一道题:"计算(2*x*x*x-3*x*x*y-2*x*y*y)-(x*x*x-2*x*y*y+y*y*y)+【-
[x(x-y)-y(x-y)+(x+y)(x-y)]÷2x其中x=2012 y=2013
(2x-y)^2-4(x-y)(x+2y)