a)\(25x^2-4=0\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b)\(x^2-6x=-9\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)c) \(\left(3x+5\right)^2-\left(2x-1\right)^2=0\Leftrightarrow\left(3x+5+2x-1\right)\left(3x+5-2x+1\right)=0\)
\(\Leftrightarrow\left(5x+4\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{5}\\x=-6\end{matrix}\right.\)
d) \(x^2-4x+3=0\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)e) \(\left(x+1\right)^2+\left(x-1\right)^2+2\left(x-1\right)\left(x+1\right)=4\)
\(\Leftrightarrow\left(x+1+x-1\right)^2=4\Leftrightarrow4x^2=4\Leftrightarrow x=\pm1\)
a) \(25x^2-4=0\)
\(\Leftrightarrow\left(5x-4\right).\left(5x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\\\5x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\\\5x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{5}\\x=-\frac{4}{5}\end{matrix}\right.\)
Vậy........
