a) \(\Leftrightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(\left(2x-2\right)+\left(3x+6\right)\right)\left(\left(2x-2\right)-\left(3x+6\right)\right)=0\)
\(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+4=0\\-x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=-8\end{cases}}}\)
b) \(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow4x+13=11\)
\(\Leftrightarrow x=-\frac{1}{2}\)
a) \(4\left(x-1\right)^2-9\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-1\right)\right]^2-\left[3\left(x+2\right)\right]^2=0\)
\(\Leftrightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)
\(\Leftrightarrow\left(2x-2+3x+6\right)\left(2x-2-3x-6\right)=0\)
\(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}5x+4=0\\-x-8=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{4}{5}\\x=-8\end{cases}}}\)
b) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x+1\right)\left(x-1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow4x+13=11\)
\(\Leftrightarrow4x=-2\)
\(\Leftrightarrow x=-\frac{2}{4}=-\frac{1}{2}\)
(Nhớ k cho mình với nhé!)
