在实数范围内分解因式:2x^2-8xy+5y^2=

最好有具体过程,谢了
2024-11-24 20:38:10
推荐回答(1个)
回答1:

2x^2-8xy+5y^2
=2x^2-8xy+8y^2-8y^2+5y^2
=2(x^2-4xy+4y^2)-3y^2
=2(x-2y)^2-3y^2
=[√2(x-2y)-y√3][√2(x-2y)+y√3]
=(x√2-2y√2-y√3)(x√2-2y√2+y√3)