Question

\(\)\(\text{Tìm x:}\)

a) \(x+\dfrac{1}{2}=2\)

b) \(75\%x-\dfrac{1}{2}=-1\dfrac{1}{4}\)

c) \(\left|x-\dfrac{2}{5}\right|+\dfrac{7}{10}=3\)

Answer 1

Answer 2

Answer 3

Answer 4

a) \(x+\dfrac{1}{2}=2\)

⇔ \(x=2-\dfrac{1}{2}\)

⇔ \(x=\dfrac{3}{2}\)

Vậy \(x=\dfrac{3}{2}\)

b) \(75\%x-\dfrac{1}{2}=-1\dfrac{1}{4}\)

⇔ \(75\%x-\dfrac{1}{2}=\dfrac{-5}{4}\)

⇔ \(\dfrac{75}{100}x=\dfrac{-5}{4}+\dfrac{1}{2}\)

⇔ \(\dfrac{3}{4}x=-\dfrac{3}{4}\)

⇔ x = -1 

Vậy x = -1 

c) \(\left|x-\dfrac{2}{5}\right|+\dfrac{7}{10}=3\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3-\dfrac{7}{10}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{23}{10}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{23}{10}\\x-\dfrac{2}{5}=\dfrac{-23}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{10}+\dfrac{2}{5}\\x=\dfrac{-23}{10}+\dfrac{2}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{10}\\x=-\dfrac{19}{10}\end{matrix}\right.\)

Answer 5

a) Ta có: \(x+\dfrac{1}{2}=2\)

nên \(x=\dfrac{3}{2}\)

Answer 6

b) Ta có: \(\dfrac{3}{4}x-\dfrac{1}{2}=\dfrac{-5}{4}\)

nên \(\dfrac{3}{4}x=\dfrac{-5}{4}+\dfrac{1}{2}=\dfrac{-5}{4}+\dfrac{2}{4}=\dfrac{-3}{4}\)

hay x=-1

Answer 7

c) Ta có: \(\left|x-\dfrac{2}{5}\right|+\dfrac{7}{10}=3\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=3-\dfrac{7}{10}=\dfrac{23}{10}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{23}{10}\\x-\dfrac{2}{5}=-\dfrac{23}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{10}\\x=-\dfrac{19}{10}\end{matrix}\right.\)