\(B=-3x^2-12x-8=-3\left(x^2+4x+4\right)+4=-3\left(x+2\right)^2+4\le4\)

Dấu \(=\)khi \(x+2=0\Leftrightarrow x=-2\).

B=-3x^2-12x-8

Ta có:-3x^2-12x-8

=-(3x^2+2.3x.2+4)+4

=-(3x+2)^2+4

Vì : (3x+2)^2 > 0

=> -(3x+2)^2 < 0

=>-(3x+2)^2+4< 4

Dấu '=' xảy ra khi (3x+2)^2=0

=>3x+2=0

=>3x=0-2

=>3x=-2

=>x=-2/3

Vậy Bmin=4 khi x=-2/3

       \(6x^2-3x-9=0\)

<=> \(6x^2+6x-9x-9=0\)

<=> \(6x.\left(x+1\right)-9.\left(x+1\right)=0\)

<=> \(\left(x+1\right).\left(6x-9\right)=0\)

<=> \(\orbr{\begin{cases}x+1=0\\6x-9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{3}{2}\end{cases}}}\)

\(6x^2-3x-9=0\)

\(6x^2-9x+6x-9=0\)

\(\left(6x^2-9x\right)+\left(6x-9\right)=0\)

\(3x\left(2x-3\right)+3\left(2x-3\right)=0\)

\(\left(2x-3\right)\left(3x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\3x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}2x=3\\3x=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}\)

Mình biết hơi muộn

\(A=x^2+2xy+6x+6y+2y^2+8\Leftrightarrow x^2+2xy+6x+6y+y^2+9-1\)

\(A=0\Rightarrow\left(x+y+3\right)^2+y^2-1=0\)

\(\Rightarrow-1\le x+y+3\le1\) .

\(\Rightarrow2012\le x+y+3+2013\le2014\)

\(\Rightarrow2012\le B\le2014\)

`Answer:`

\(A+\left(6x^2y+2xy+4y\right)=7x^2y-y\)

\(\Leftrightarrow A=\left(7x^2y-y\right)-\left(6x^2y+2xy+4y\right)\)

\(\Leftrightarrow A=\left(7x^2y-6x^2y\right)+\left(-y-4y\right)-2xy\)

\(\Leftrightarrow A=x^2y-5y-2xy\)

giải giúp mik nha please

\(A+\left(6x^2y+2xy+4y\right)=7x^2y-y\)

\(A=7x^2y-y-\left(6x^2y+2xy+4y\right)\)

\(A=7x^2y-y-6x^2y-2xy-4y\)

\(A=\left(7x^2y-6x^2y\right)+\left(-y-4y\right)-2xy\)

\(A=x^2y-5y-2xy\)

A+6x^2y+2xy+4y=7x^2y-y

A=7x^2y-y-6x^2y-2xy-4y

A=x^2y-5y-2xy

A=x^2y-5y-2xy

cảm ơn bn nha :))

là m hả T.kim !!!!

Đặt \(f\left(x\right)=x^3-2x^2-6x+a\)

Gọi thương của \(f\left(x\right):\left(x-2\right)\)là \(P\left(x\right)\)

\(\Rightarrow f\left(x\right)=P\left(x\right).\left(x-2\right)\)

Thay \(x=2\)ta có: 

\(8-8-12+a=0\)

\(\Rightarrow a=12\)

Vậy \(a=2\)là giá trị cần tìm