Question

Answer 1

Bài 18"

\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{10}\right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{15-10-3}{30}=\dfrac{2}{30}=\dfrac{1}{15}\)

\(\dfrac{1}{12}-\left(-\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{1+2+3}{12}=\dfrac{6}{12}=\dfrac{1}{2}\)

\(\dfrac{1}{2}-\dfrac{-1}{3}+\dfrac{1}{23}+\dfrac{1}{6}=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{23}+\dfrac{1}{6}=\dfrac{3+2+1}{6}+\dfrac{1}{23}=1+\dfrac{1}{23}=\dfrac{24}{23}\)

Answer 2

giúp mik với

Answer 3

nhỏ quá

 

 

 

 

Answer 4

Bài 18':

a) Ta có: \(x-\dfrac{1}{15}=\dfrac{1}{10}\)

nên \(x=\dfrac{1}{10}+\dfrac{1}{15}\)

hay \(x=\dfrac{1}{6}\)

Vậy: \(x=\dfrac{1}{6}\)

b) Ta có: \(\dfrac{-2}{15}-x=\dfrac{-3}{10}\)

nên \(x=\dfrac{-2}{15}-\dfrac{-3}{10}\)

hay \(x=\dfrac{1}{6}\)

Vậy: \(x=\dfrac{1}{6}\)

Answer 5

Bài 18':

a) Ta có: \(\dfrac{-2}{3}x=\dfrac{4}{15}\)

\(\Leftrightarrow x=\dfrac{4}{15}:\dfrac{-2}{3}=\dfrac{4}{15}\cdot\dfrac{-3}{2}=\dfrac{-12}{30}\)

hay \(x=-\dfrac{2}{5}\)

Vậy: \(x=-\dfrac{2}{5}\)

c) Ta có: \(\dfrac{-21}{13}x+\dfrac{1}{3}=\dfrac{-2}{3}\)

\(\Leftrightarrow x\cdot\dfrac{-21}{13}=-1\)

hay \(x=\dfrac{13}{21}\)

Vậy: \(x=\dfrac{13}{21}\)