a) \(\left(x-3\right)^2-x\left(x+2\right)=1\Leftrightarrow x^2-6x+9-x^2-2x=1\Leftrightarrow-8x=-8\Leftrightarrow x=1\)
b) \(\left(3x+2\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow9x^2+12x+4-4x^2+20x-25=0\Leftrightarrow5x^2+32x-21=0\Leftrightarrow5\left(x-\dfrac{3}{5}\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
c) \(x^2-4=5\left(2-x\right)\Leftrightarrow x^2-4+5x-10=0\Leftrightarrow x^2+5x-14=0\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
a)\(\left(x-3\right)^2-x\left(x+2\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-2x=1\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\)
Vậy x=1
b)\(\left(3x+2\right)^2-\left(2x-5\right)^2=0\\ \Leftrightarrow\left(3x+2+2x-5\right)\left(3x+2-2x+5\right)=0\\ \Leftrightarrow\left(5x-3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x-3=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-7\end{matrix}\right.\\\)
a: Ta có: \(\left(x-3\right)^2-x\left(x+2\right)=1\)
\(\Leftrightarrow x^2-6x+9-x^2-2x=1\)
\(\Leftrightarrow-8x=-8\)
hay x=1
b: Ta có: \(\left(3x+2\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(3x+2-2x+5\right)\left(3x+2+2x-5\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=\dfrac{3}{5}\end{matrix}\right.\)
