Question

1. Tìm x

a. 2.(4-3x)+2x=5(2x-3)

b. \(\dfrac{1}{2}-\left(2x-\dfrac{1}{3}\right)^2=\dfrac{7}{18}\)

Answer 1

Answer 2

a: ta có: \(2\left(4-3x\right)+2x=5\left(2x-3\right)\)

\(\Leftrightarrow8-6x+2x-10x+15=0\)

\(\Leftrightarrow-14x=-23\)

hay \(x=\dfrac{23}{14}\)

b: Ta có: \(\dfrac{1}{2}-\left(2x-\dfrac{1}{3}\right)^2=\dfrac{7}{18}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{3}\right)^2=\dfrac{1}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{3}=\dfrac{1}{3}\\2x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{2}{3}\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=0\end{matrix}\right.\)