Question

giải giúp với mn!

Answer 1

BT2:

\(1)\dfrac{4}{5}-\dfrac{1}{5}:\left(x+\dfrac{3}{10}\right)=1\dfrac{3}{5}\\ \dfrac{4}{5}-\dfrac{1}{5}:\left(x+\dfrac{3}{10}\right)=\dfrac{8}{5}\\\dfrac{1}{5}:\left(x+\dfrac{3}{10}\right)=\dfrac{4}{5}-\dfrac{8}{5}=-\dfrac{4}{5}\\ x+\dfrac{3}{10}=\dfrac{1}{5}:-\dfrac{4}{5}=-\dfrac{1}{4}\\ x=-\dfrac{1}{4}-\dfrac{3}{10}\\ x=-\dfrac{11}{20}\\ 2)-\dfrac{22}{15}x+\dfrac{1}{3}=\left|-\dfrac{2}{3}+\dfrac{1}{5}\right|\\ -\dfrac{22}{15}x+\dfrac{1}{3}=-\dfrac{7}{15}\\ -\dfrac{22}{15} x=-\dfrac{7}{15}-\dfrac{1}{3}\\ -\dfrac{22}{15}x=-\dfrac{4}{5}\\ x=-\dfrac{4}{5}:-\dfrac{22}{15}\\ x=\dfrac{6}{11}\\ 3)1,75-\left(\dfrac{3}{4}-x\right)\cdot0,5=-\dfrac{1}{6}\\ \dfrac{7}{4}-\dfrac{1}{2}\left(\dfrac{3}{4}-x\right)=-\dfrac{1}{6}\\ \dfrac{1}{2}\left(\dfrac{3}{4}-x\right)=\dfrac{7}{4}+\dfrac{1}{6}\\ \dfrac{1}{2}\left(\dfrac{3}{4}-x\right)=\dfrac{23}{12}\\ \dfrac{3}{4}-x=\dfrac{23}{12}:\dfrac{1}{2}=\dfrac{23}{6}\\ x=\dfrac{3}{4}-\dfrac{23}{6}\\ x=-\dfrac{37}{12}\)

Answer 2

Bài 1:

1: \(\left(\dfrac{12}{35}-\dfrac{6}{7}+\dfrac{18}{14}\right):\dfrac{6}{-7}-\dfrac{-2}{5}\)

\(=\left(\dfrac{12}{35}-\dfrac{30}{35}+\dfrac{45}{35}\right)\cdot\dfrac{-7}{6}+\dfrac{2}{5}\)

\(=\dfrac{27}{35}\cdot\dfrac{-7}{6}+\dfrac{2}{5}=\dfrac{-9}{10}+\dfrac{2}{5}=\dfrac{-9}{10}+\dfrac{4}{10}=-\dfrac{5}{10}=-\dfrac{1}{2}\)

2: \(\left(-\dfrac{2}{3}\right)^2\cdot0,5-\left|-\dfrac{1}{2}\right|\cdot\sqrt{16}\)

\(=\dfrac{4}{9}\cdot\dfrac{1}{2}-\dfrac{1}{2}\cdot4=\dfrac{2}{9}-2=\dfrac{2}{9}-\dfrac{18}{9}=-\dfrac{16}{9}\)

3: \(\dfrac{7}{12}\cdot3\dfrac{2}{5}-\dfrac{7}{12}\cdot9\dfrac{2}{5}\)

\(=\dfrac{7}{12}\left(3+\dfrac{2}{5}-9-\dfrac{2}{5}\right)\)

\(=\dfrac{7}{12}\cdot\left(-6\right)=-\dfrac{7}{2}\)

4: \(-6\dfrac{11}{14}-\left(1\dfrac{3}{12}-3\dfrac{11}{14}\right)\)

\(=-6-\dfrac{11}{14}-1-\dfrac{1}{4}+3+\dfrac{11}{14}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)