15 phút nữa đưa ra lời giải rồi đợi mọi người bấm à

\(\left(x+y\right)^6+\left(x-y\right)^6=\left[\left(x+y\right)^2\right]^3+\left[\left(x-y\right)^2\right]^3\) chia hết cho \(\left(x+y\right)^2+\left(x-y\right)^2\) tức là chia hết cho \(2.\left(x^2+y^2\right)\) do đó chia hết cho \(x^2+y^2\)

Ta có : \(\left(x+y\right)^2=x^2+2xy+y^2.\)

\(\Leftrightarrow\left(x+y\right)^6=\left(x+y\right)^2.\left(x+y\right)^2.\left(x+y\right)^2\) .

\(\Leftrightarrow\left(x+y\right)^6=x^6+8x^3y^3+y^6\) .

\(\Leftrightarrow\left(x-y\right)^6=x^6-8x^3y^3+y^6\).

\(\Leftrightarrow\left(x+y\right)^6+\left(x-y\right)^6=\left(x^6+x^6\right)+\left(8x^3y^3-8x^3y^3\right)+\left(x^6+x^6\right)\).

\(\Leftrightarrow\left(x+y\right)^6+\left(x-y\right)^6=2.x^6+2.y^6=2\left(x^6+y^6\right)=2.\left(\left(x^2+y^2\right).\left(x^2+y^2\right):2\right).\)

\(\left(x+y\right)^6+\left(x-y\right)^6\)

\(=\left[\left(x+y\right)^2\right]^3+\left[\left(x-y\right)^2\right]^3\)

\(=\left[\left(x+y\right)^2+\left(x-y\right)^2\right]\left[\left(x+y\right)^4-\left(x+y\right)\left(x-y\right)+\left(x+y\right)^4\right]\)

\(=\left(2x^2+2y^2\right)\left[\left(x+y\right)^4-\left(x+y\right)\left(x-y\right)+\left(x+y\right)^4\right]\)

\(=2\left(x^2+y^2\right)\left[\left(x+y\right)^4-\left(x+y\right)\left(x-y\right)+\left(x+y\right)^4\right]\)

Ta có: (x²+y²) \(⋮\) x² + y²

=> \(2\left(x^2+y^2\right)\left[\left(x+y\right)^4-\left(x+y\right)\left(x-y\right)+\left(x+y\right)^4\right]\) \(⋮\) \(x^2+y^2\)

Lời giải:

Đặt \((x+y)^2=a; (x-y)^2=b\)

\(\Rightarrow a+b=2(x^2+y^2)\)

Khi đó:

\((x+y)^6+(x-y)^6=a^3+b^3=(a+b)(a^2-ab+b^2)=2(x^2+y^2)(a^2-ab+b^2)\vdots x^2+y^2\)

Ta có đpcm.

\(\left(\left(x+y\right)^2\right)^3+\left(\left(x-y\right)^2\right)^3\)

\(=\left(\left(x+y\right)^2+\left(x-y\right)^2\right)\left(\left(x+y\right)^4-\left(x^2-y^2\right)^2+\left(x-y\right)^4\right)\)

\(=\left(2x^2+2y^2\right)\left(\left(x+y\right)^4-\left(x^2-y^2\right)^2+\left(x-y\right)^4\right)\)

\(=2\left(x^2+y^2\right)\left(\left(x+y\right)^4-\left(x^2-y^2\right)^2+\left(x-y\right)^4\right)⋮\left(x^2+y^2\right)\)

\(\left(x+y\right)^6+\left(x-y\right)^6\)

\(=\left[\left(x+y\right)^2\right]^3+\left[\left(x-y\right)^2\right]^3\)

\(=\left[\left(x+y\right)^2+\left(x-y\right)^2\right]\left(...\right)\)

\(=\left(x^2+2xy+y^2+x^2-2xy+y^2\right)\left(...\right)\)

\(=\left(2x^2+2y^2\right)\left(...\right)\)

\(=2\left(x^2+y^2\right)\left(...\right)⋮x^2+y^2\left(đpcm\right)\)

1) A=\(\left(x+y\right)^6+\left(x-y\right)^6=\left[\left(x+y\right)^2+\left(x-y\right)^2\right]\left[binh-phuong-thieu\right]\)

                                             \(=2\left(x^2+y^2\right)\left[binh-phuong-thieu..\right]\)=> A chia hết cho x²+y²

2)  gọi dư của phép chia là ax+b

 ta có f(1) = a+b =51

         f(-1) = -a+b =1 

=> b =26 ; a =25

Vậy dư là : 25x + 26

(x+y)(y+z)(z+x)-2xyz

⇒(x+y+z)-z(x+y+z)-x(x+y+z)-y-2xyz

⇒(x+y+z)nhân-(x+y+z)-2xyz

⇒6(-6)-2xyz⋮6

⇒(x+y)(y+z)(z+x)-2xyz⋮6