Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Có: \(\left(x+y\right)^4+x^4+y^4\)
\(=\left(x+y\right)^4+\left(x^2+y^2\right)^2-2x^2y^2\)
\(=\left[\left(x+y\right)^4-x^2y^2\right]+\left[\left(x^2+y^2\right)^2-x^2y^2\right]\)
\(=\left[\left(x^2+y^2+2xy\right)^2-\left(xy\right)^2\right]+\left[\left(x^2+y^2\right)^2-\left(xy\right)^2\right]\)
\(=\left(x^2+y^2+xy\right)\left(x^2+y^2+3xy\right)+\left(x^2+y^2+xy\right)\left(x^2+y^2-xy\right)\)
\(=2\left(x^2+y^2+xy\right)\left(x^2+y^2+xy\right)=2\left(x^2+y^2+xy\right)^2\)
x4y4+64=x4y4+16x2y2+64-16x2y2
=(x2y2+8)2-16x2y2
=(x2y2-4xy+8)(x2y2+4xy+8)
x^4 - y^4
= (x^2 - y^2)(x^2 + y^2)
= (x - y)(x + y)(x^2 + y^2)
\(x^4+y^4+\left(x+y\right)^4\)
\(=x^4+y^4+\left(x^2+2xy+y^2\right)^2\)
\(=x^4+y^4+x^4+6x^2y^2+y^4+4x^3y+4xy^3\)
\(=2.\left(x^2+y^2\right)^2+4xy\left(x^2+y^2\right)+2x^2y^2\)
\(=2.\left(x^2+y^2\right)\left(x^2+y^2+2xy\right)+2x^2y^2\)
\(=2.\left[\left(x^2+y^2\right)\left(x+y\right)^2+x^2y^2\right]\)
Sai thì thôi nhé~
\(x^4+y^4+\left(x+y\right)^4\)
\(=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
\(=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\)
\(=2\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)\)
\(=2\left[\left(x^4+2x^3y+x^2y^2\right)+2\left(x^2+xy\right)y^2+y^4\right]\)
\(=2\left[\left(x^2+xy\right)^2+2\left(x^2+xy\right)y^2+\left(y^2\right)^2\right]\)
\(=2\left(x^2+xy+y^2\right)^2\)
\(x^4.y^4+4\)
\(=\left(x^4y^4-2x^3y^3+2x^2y^2\right)+\left(2x^3y^3-4x^2y^2+4xy\right)+\left(2x^2y^2-4xy+4\right)\)
\(=x^2y^2\left(x^2y^2-2xy+2\right)+2xy\left(x^2y^2-2xy+2\right)+2\left(x^2y^2-2xy+2\right)\)
= (x2y2 + 2xy + 2)(x2y2 - 2xy + 2)
Dùng cách này cho nhanh :v
Đặt xy = t cho dễ nhìn. \(t^4+4=\left(t^4+2t^2.2+4\right)-\left(2t\right)^2\)
\(=\left(t^2+2\right)^2-\left(2t\right)^2=\left(t^2-2t+2\right)\left(t^2+2t+2\right)\)
\(=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)