f) \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
h) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
i) \(2x\left(x-2\right)-\left(2-x\right)^2=0\)
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=\pm2\)
k) \(\left(1-x\right)^2-1+x=0\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
l) \(\left(x-3\right)^3+3-x=0\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\Leftrightarrow x=4\end{matrix}\right.\)
m) \(x+6x^2=0\)
\(\Leftrightarrow x\left(1+6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{6}\end{matrix}\right.\)
n) \(\left(x+1\right)=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
f ) \(4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy ...
h ) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Vậy ...
I ) \(2x\left(x-2\right)-\left(2-x\right)^2=0\)
\(\Leftrightarrow-2x\left(2-x\right)-\left(2-x\right)^2=0\)
\(\Leftrightarrow\left(-2x-2+x\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left(-2-x\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2-x=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Vậy ...
K ) \(\left(1-x\right)^2-1+x=0\)
\(\Leftrightarrow\left(1-x\right)^2-\left(1-x\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(1-x\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
Vậy ...
i ) \(\left(x-3\right)^3+3-x=0\)
\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\left(x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x-3\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\\x=2\end{matrix}\right.\)
Vậy ...
m ) \(x+6x^2=0\)
\(\Leftrightarrow x\left(1+6x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\1+6x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\6x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{6}\end{matrix}\right.\)
Vậy ...
n ) \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vậy ...
