2x^2-3xy+y^2=0怎么因式分解
来源:百度知道 编辑:UC知道 时间:2024/05/26 19:41:22
麻烦给我具体过程,越详尽越好~~~
十字相乘法,把y看作常数...
2=1*2,写成1 -y
2 -y
对角相乘应满足3xy……
故分解成(2x-y)(x-y)=0
够详尽了吧....
也可用提公因式法
2x^2-3xy+y^2
=2x^2-2xy+y^2-xy
=x^2+x^2-2xy+y^2-xy
=x^2+(x-y)^2-xy
=x(x-y)+(x-y)^2
2x^2-3xy+y^2
=(x-y)^2+x^2-xy
=(x-y)^2+x(x-y)
=(x-y)(x-y+x)
=(x-y)(2x-y)
2x^2-3xy+y^2
=2x^2-2xy+y^2-xy(拆-3xy)
=x^2+x^2-2xy+y^2-xy(拆2x^2)
=x^2+(x-y)^2-xy(完全平方公式合并x^2-2xy+y^2)
=x(x-y)+(x-y)^2(提取x^2和xy的公因式x)
=(x-y)(2x-y)(提取(x-y))
(2x-y)(x-y)=0
2x=y x=y
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