a, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)
ĐKXĐ: \(\forall x\in R\)
\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=7 \)
\(\Leftrightarrow x^2+6x+9+x^2-4-2x^2+4x-2=7\)
\(\Leftrightarrow10x+3=7\)
\(\Leftrightarrow10x=4\Leftrightarrow x=\frac{4}{10}=\frac{2}{5}\)
Vậy \(x=\frac{2}{5}\)
b, \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
ĐKXĐ: \(\forall x\in R\)
\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy x = -2.
c,\(\left(3x-2\right)^2-\left(2x-1\right)^2=0\)
ĐKXĐ: \(\forall x\in R\)
\(\Leftrightarrow\left(3x-2+2x-1\right)\left(3x-2-2x+1\right)=0\\ \Leftrightarrow\left(5x-3\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{5}\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{5};1\right\}\)
d, \(x^2-4x+3=0\)
ĐKXĐ: \(\forall x\in R\)
\(\Leftrightarrow\left(x^2-x\right)-\left(3x-3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{1;3\right\}\)