a.
\(3-2x=3\left(x+1\right)-x-2\)
\(\Leftrightarrow3-2x=3x+3-x-2\)
\(\Leftrightarrow-2x-3x+x=3-2-3\)
\(\Leftrightarrow-4x=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy.........
b.
\(\left(3x+2\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\4x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy........
c.
\(\left(x+2\right)\left(3x+1\right)+x^2=4\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+x^2-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy...........
a.3-2x=3(x+1)-x-2
\(\Leftrightarrow3-2x=3x+3-x-2\)
\(\Leftrightarrow-2x+x-3x=-3+3-2\)
\(\Leftrightarrow-4x=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy S={1/2}
b. (3x+2)(4x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\4x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy S=\(\left\{-\dfrac{2}{3};\dfrac{5}{4}\right\}\)
c.\(\left(x+2\right)\left(3x+1\right)+x^2=4\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+x^2-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy S=\(\left\{-2;\dfrac{1}{4}\right\}\)
