a,x^2+2xy+7x+7y+y^2+10a

=(x^2+2xy+y^2)+(7x+7y)+10

=(x+y)^2+7(x+y)+10

=(x+y)(x+y+tx7)+10

Đặt x+y=t ,có :

t(t+7)+10

=t^2+7t+10

=t^2+2t+5t+10

=t(t+2)+5(t+2)

=(t+2)(t+5)

=(x+y+2)(x+y+5)

để đa thức B=0 thì :

x^2-2x=0

x(x-2)=0

\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy no của đa thức là x=0 hoặc x=2

B(x)  = x^2 –  2x        

x^2 - 2x =0

=> x(x-2) = 0 

=> \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

     C(x) = x^2 – 5x – 6

=>  x^2 - 6x + x - 6 = 0 

=>   (x^2 - 6x) + (x - 6) = 0 

=> x( x - 6) + ( x- 6) = 0 

=> (x+1 )( x - 6 ) =0

=> \(\orbr{\begin{cases}x=-1\\x=6\end{cases}}\)

Đây nha cậu 

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bằng 63

\(C=4x^2+9y^2+4x-9y+3\)

\(=\left(4x^2+4x+1\right)+\left(9y^2-9y+\frac{9}{4}\right)+3-1-\frac{9}{4}\)

\(=\left(2x+1\right)^2+\left(3y-\frac{3}{2}\right)^2-\frac{1}{4}\)

Mà \(\left(2x+1\right)^2+\left(3y-\frac{3}{2}\right)^2\ge0\Rightarrow\left(2x+1\right)^2+\left(3y-\frac{3}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy \(C_{Min}=-\frac{1}{4}\)khi và chỉ khi \(\hept{\begin{cases}\left(2x+1\right)^2=0\\\left(3y-\frac{3}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+1=0\\3y-\frac{3}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}}\)

\(D=2x^2+y^2+2xy-10x+2y+2023\)

\(=\left(x^2+2xy+y^2\right)+x^2-10x+2y+2023\)

\(=\left(x+y\right)^2+2x+2y+\left(x^2-12x+36\right)+2023-36\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right)+1\right]+\left(x-6\right)^2+1987-1\)

\(=\left(x+y+1\right)^2+\left(x-6\right)^2+1986\)

Mà \(\left(x+y+1\right)^2+\left(x-6\right)^2\ge0\Rightarrow\left(x+y+1\right)^2+\left(x-6\right)^2+1986\ge1986\)

Vậy \(D_{Min}=1986\)khi và chỉ khi \(\hept{\begin{cases}\left(x+y+1\right)^2=0\\\left(x-6\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+y+1=0\\x-6=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=6\\y=-7\end{cases}}}\)

Đặt `a+b=x, b+c=y, c+a=z`

`->x+y+z=2 (a+b+c)`

`(a+b)^3 +(b+c)^3 + (c+a)^3 - 8 (a+b+c)^3`

`= x^3 + y^3 + z^3 - 2^3 (a+b+c)^3`

`=x^3 +y^3 +z^3 - [2 (a+b+c)]^3`

`=x^3 +y^3+z^3 - (x+y+z)^3`

`= x^3 + y^3 +z^3 - [x^3 +y^3 +z^3 + 3 (x+y) (y+z) (x+z)]`

`= -3 (x+y)(y+z)(x+z)`

`= -3 (2b + a+c) (2c+a+b) (2a +b+c)`

Đặt : \(\hept{\begin{cases}a+b=x\\b+c=y\\c+a=z\end{cases}}\Rightarrow\left(a+b\right)^3+\left(b+c\right)^3+\left(a+c\right)^3-8\left(a+b+c\right)^3=x^3+y^3+z^3-\left(x+y+z\right)^3\)

\(=x^3+y^3+z^3-\left(x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\right)=-3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(=-3\left(2a+b+c\right)\left(a+2b+c\right)\left(a+b+2c\right)\)