\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(x+2+3x-6\right)\left(x+2-3x+6\right)=0\)
\(\Leftrightarrow\left(4x-4\right)\left(-2x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\-2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)
Vậy ...
a) \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x+7\right)^2-\left[3\left(x+2\right)\right]^2=0\)
\(\Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(2x+7-3x-6\right)\left(2x+7+3x+6\right)=0\)
\(\Leftrightarrow\left(1-x\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-x=0\\5x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=-13\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{5}\end{matrix}\right.\)
b)\(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-9\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left[3\left(x-2\right)\right]^2=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(3x-6\right)^2=0\)
\(\Leftrightarrow\left(x+2-3x+6\right)\left(x+2+3x-6\right)=0\)
\(\Leftrightarrow\left(8-2x\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8-2x=0\\4x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(4-x\right)=0\\4\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow4x^2+28x+49=9x^2+36x+36\)
\(\Leftrightarrow5x^2+8x-13=0\)
\(\Leftrightarrow5x^2-5x+13x-13=0\)
\(\Leftrightarrow5x\left(x-1\right)+13\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{13}{5}\end{matrix}\right.\)
Vậy ...
