Question

1.Tìm x

a) \(1\dfrac{4}{5}=-0,15-x\)

b) \(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\)

c) \(\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\)

Answer 1

\(a,\Leftrightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\\ \Leftrightarrow-\dfrac{9}{5}=x+\dfrac{3}{20}\\ \Leftrightarrow x=-\dfrac{9}{5}-\dfrac{3}{20}=-\dfrac{39}{20}\\ b,\Leftrightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Leftrightarrow x=\dfrac{41}{35}\\ c,\Leftrightarrow\dfrac{3}{8}-\dfrac{1}{5}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Leftrightarrow1-x=\dfrac{2}{5}\\ \Leftrightarrow x=1-\dfrac{2}{5}=\dfrac{3}{5}\)

Answer 2

Giúp mình với 

Answer 3

a: Ta có: \(-x-\dfrac{3}{20}=\dfrac{9}{5}\)

\(\Leftrightarrow x=-\dfrac{9}{5}+\dfrac{3}{20}=\dfrac{-36+3}{20}=\dfrac{-33}{20}\)

b: Ta có: \(\dfrac{-4}{7}-x=\dfrac{3}{5}-2x\)

\(\Leftrightarrow x=\dfrac{3}{5}+\dfrac{4}{7}=\dfrac{41}{35}\)