Question

 giúp bài 3, câu c,f,g,l,k bài 2 

Answer 1

Bài 2:

c: \(\left(\dfrac{3}{2}+x\right):1\dfrac{2}{5}=\dfrac{1}{2}\cdot\dfrac{3}{5}+0,2\)

=>\(\left(x+\dfrac{3}{2}\right):\dfrac{7}{5}=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{5}{10}=\dfrac{1}{2}\)

=>\(x+\dfrac{3}{2}=\dfrac{1}{2}\cdot\dfrac{7}{5}=\dfrac{7}{10}\)

=>\(x=\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{7}{10}-\dfrac{15}{10}=-\dfrac{8}{10}=-\dfrac{4}{5}\)

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

=>\(-\dfrac{7}{5}-\dfrac{2}{3}-x=\dfrac{3}{10}\)

=>\(x=-\dfrac{7}{5}-\dfrac{2}{3}-\dfrac{3}{10}=\dfrac{-42-20-9}{30}=\dfrac{-71}{30}\)

g: \(\left|x-\dfrac{1}{6}\right|-\dfrac{7}{12}=\dfrac{1}{4}\)

=>\(\left|x-\dfrac{1}{6}\right|=\dfrac{1}{4}+\dfrac{7}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)

=>\(\left[{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{5}{6}\\x-\dfrac{1}{6}=-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{6}=1\\x=-\dfrac{5}{6}+1=-\dfrac{4}{6}=-\dfrac{2}{3}\end{matrix}\right.\)

l: \(x:\left(\dfrac{1}{5}-\dfrac{7}{10}\right)=-2+\left(-1\dfrac{2}{3}\right)\)

=>\(x:\left(\dfrac{2}{10}-\dfrac{7}{10}\right)=-2+\dfrac{-5}{3}=\dfrac{-11}{3}\)

=>\(x:\dfrac{-1}{2}=-\dfrac{11}{3}\)

=>\(x=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)

k: \(x:\left(\dfrac{1}{7}-\dfrac{3}{14}\right)=-3+\left(-1\dfrac{2}{3}\right)\)

=>\(x:\left(\dfrac{2}{14}-\dfrac{3}{14}\right)=-3-\dfrac{5}{3}=\dfrac{-14}{3}\)

=>\(x:\dfrac{-1}{14}=\dfrac{-14}{3}\)

=>\(x=\dfrac{14}{3}\cdot\dfrac{1}{14}=\dfrac{1}{3}\)

Bài 3:

a: \(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)

\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

b: \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)

\(=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)

\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{5-2}{20}=\dfrac{3}{20}\)