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.

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

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

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

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

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

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

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

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

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

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

(x+y)^6+(x-y)^6=[(x+y)²]³+[(x-y)²]³=[(x+y)²+(x-y)²][(x+y)⁴-(x+y)(x-y)+(x-y)⁴]

=(2x²+2y²)[(x+y)⁴-(x+y)(x-y)+(x-y)⁴]

=2.(x²+y²)[(x+y)⁴-(x+y)(x-y)+(x-y)⁴] chia hết cho đa thức x²+y²

=> điều phải chứng minh

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

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

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

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

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

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

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

Bài 1:

a) x^3 + 2x^2 + x = x.(x^2+2x+1) = x.(x+1)^2

b) xy + y^2 - x - y

= y.(x+y) - (x+y)

= (x+y).(y-1)

Bài 2:

a) 3x.(x^2-4) = 0

3x.(x+2).(x-2) = 0

=> x = 0; x = -2; x = 2

b) x^2 - 4x + 3 = 0

x^2 - x - 3x + 3 = 0

x.(x-1) - 3.(x-1) = 0

(x-1).(x-3) = 0

=> x = 1; x = 3