\(7\left(x+1\right)+3x=27\)
\(7x+7+3x=27\)
\(10x=20\)
\(x=2\)
\(\left(x+2\right)\left(3-2x\right)+x=2x^2-3\)
\(3x+6-2x^2-4x+x=2x^2-3\)
\(-2x^2-2x^2=-3-6\)
\(-4x^2\)=\(=-9\)
\(x^2=\dfrac{9}{4}\)
\(=>x\in\left\{\dfrac{3}{2};\dfrac{-3}{2}\right\}\)
\(7\left(x+1\right)+3x=27\\ \Leftrightarrow7x+7+3x=27\\ \Leftrightarrow10x=20\\ \Leftrightarrow x=2\)
Vậy x = 2
\(\left(x+2\right)\left(3-2x\right)+x=2x^2-3\\ \Leftrightarrow3x-4x-2x^2+6+x=2x^2-3\\ \Leftrightarrow-2x^2+6=2x^2-3\\ \Leftrightarrow4x^2=9\\ \Leftrightarrow x^2=\dfrac{9}{4}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\)
