\(\hept{\begin{cases}x^2=8x+y\\y^2=8y+x\end{cases}\Leftrightarrow\left(x-y\right)\left(x+y\right)=7\left(x-y\right)vi.x\ne}y\Leftrightarrow x+y=7\)

\(\Leftrightarrow x^2+y^2=8\left(x+y\right)+\left(y+x\right)\Rightarrow DS=9\cdot7=63\)

x² = 8x + y

y² = x + 8y

=> x² - y² = (8x + y) - (x + 8y)

(x - y)(x + y) = 8x + y - x - 8y

(x - y)(x + y) = 7x - 7y

(x - y)(x + y) = 7(x - y)

x + y = 7

x² = 8x + y

y² = x + 8y

=> x² + y² = 8x + y + x + 8y = 9x + 9y = 9(x + y) = 9 . 7 = 63

ĐS: 63

\(\begin{matrix}x=3,y=-3\\x=-3,y=3\end{matrix}\)

Mik nghĩ vậy, không chắc lắm,

Đúng thì tick cho mình nha

\(1,4x^4+4x^2y^2-8y^4\)

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

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

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

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

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

\(2,12x^2y-18xy^2-30y^3\)

\(=6y\left(2x^2-3xy-5y^2\right)\)

\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)

\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)

\(=6y\left(x+y\right)\left(2x-5y\right)\)

Ta có: C + D = 8

<=> 2x^3 - 6x^2 + 8x + 2y^3 - 6y^2 + 8y = 8

<=> x^3 - 3x^2 + 4x + y^3 - 3y^2 + 4y = 4

<=> ( x^3 - 3x^2 + 3x - 1 ) + ( y^3 - 3y^2 + 3y - 1 ) + ( x + y - 2 ) = 0

<=> ( x - 1 )^3 + ( y - 1 )^3 + ( x + y - 2 ) = 0

<=> ( x - 1 + y - 1 ) . ( ( x - 1 )^2 - ( x - 1 )( y - 1 ) + ( y - 1 )^2 ) + ( x + y - 2 ) = 0

<=> ( x + y - 2 ) . A + ( x + y - 2 ) = 0

<=> ( x + y - 2 ) . ( A + 1 ) = 0

<=> x + y - 2 = 0               ( vì A lớn hơn hoặc = 0 nên A + 1 > 0 ) 

<=> x + y = 2.

Vậy x + y = 2.

Minh rảnh nhỉ??? Tự hỏi tự trả lời :))

a)  \(A=x^2+8x=x^2+8x+16-16=\left(x+4\right)^2-16\ge-16\)

Vậy MIN \(A=-16\)khi  \(x=-4\)

b)  \(B=x^2-4x+y^2-8y+6=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(=\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\)

Vậy MIN   \(B=-14\) khi  \(x=2;\)\(y=4\)