a) x Î{0;3}.

b) xÎ{0;-9}.

c) x Î{-l; 11}.

d) x = 13.

3) \(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow\left(x-4\right)\left(x+x-4\right)=0\Leftrightarrow2\left(x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

4x.(x+1)-8(x+1)=0

(4x-8)(x+1)=0

suy ra x=2 hoặc x=-1

1) \(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow4x^2+4x=8x+8\Leftrightarrow4x^2-4x-8=0\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

a) -3(x-4)+5(x-1)=-7

=>-3x+12+5x-5=-7

=>2x+7=-7

=>2x=-14=>x=-7

b) -4./x-8/+12=0

=>/x-8/=3

=>x-8=3 hoặc -3

(tự tính) 

Ta có : \(\left(8-x\right).\left(x-1\right)>0\)

\(\Leftrightarrow\hept{\begin{cases}8-x>0\\x-1>0\end{cases}\Rightarrow}\hept{\begin{cases}x< 8\\x>1\end{cases}}\left(TM\right)\)

hoặc : \(\hept{\begin{cases}8-x< 0\\x-1< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>8\\x< 1\end{cases}}\left(KTM\right)\)

\(\Rightarrow x\in\left\{2;3;4;5;6;7\right\}\)thì \(\left(8-x\right).\left(x-1\right)>0\)

\(a,\Leftrightarrow x^2-x-x^2+6x+16=1\\ \Leftrightarrow5x=-15\Leftrightarrow x=-3\\ b,\Leftrightarrow2x\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)