Question

tìm x

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

\(x+\dfrac{7}{8}=\dfrac{3}{4}\)

\(\dfrac{1}{2}.x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\dfrac{1}{2}-\dfrac{5}{6}:x=\dfrac{2}{3}\)

 

 

Answer 1

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

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

\(x=\dfrac{5}{4}\)

\(x+\dfrac{7}{8}=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}-\dfrac{7}{8}\)

\(x=\dfrac{-1}{8}\)

\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)

\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)

\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)

\(x=\dfrac{-1}{2}\)

Câu D ko bt