Bài 2: Tìm x, biết:

a) \(\dfrac{2}{3}x+\dfrac{1}{2}=\dfrac{1}{10}\\ =>\dfrac{2}{3}x=\dfrac{1}{10}-\dfrac{1}{2}=-\dfrac{2}{5}\\ =>x=\dfrac{\dfrac{-2}{5}}{\dfrac{2}{3}}=-\dfrac{3}{5}\)

b) \(\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{7}{10}\\ =>\dfrac{2}{3}x=\dfrac{7}{10}-\dfrac{1}{5}=\dfrac{1}{2}\\ =>x=\dfrac{\dfrac{1}{2}}{\dfrac{2}{3}}=\dfrac{3}{4}\)

c) \(\left(3\dfrac{4}{5}-2x\right).1\dfrac{1}{3}=5\dfrac{5}{7}\\ < =>\left(\dfrac{19}{5}-2x\right).\dfrac{4}{3}=\dfrac{40}{7}\\ =>\dfrac{19}{5}-2x=\dfrac{\dfrac{40}{7}}{\dfrac{4}{3}}=\dfrac{30}{7}\\ =>2x=\dfrac{19}{5}-\dfrac{30}{7}=-\dfrac{17}{35}\\ =>x=\dfrac{-\dfrac{17}{35}}{2}=-\dfrac{17}{70}\)

d) \(\dfrac{x}{7}=\dfrac{6}{-21}\\ =>x=\dfrac{6.7}{-21}=-2\)

\(a,\dfrac{2}{3}.x+\dfrac{1}{2}=\dfrac{1}{10}\)

\(\Rightarrow\dfrac{2}{3}.x=\dfrac{1}{10}-\dfrac{1}{2}=\dfrac{-2}{5}\)

\(\Rightarrow x=\dfrac{-2}{5}:\dfrac{2}{3}=\dfrac{-2}{5}.\dfrac{3}{2}=\dfrac{3}{5}\)

\(b,\dfrac{2}{3}.x+\dfrac{1}{5}=\dfrac{7}{10}\)

\(\Rightarrow\dfrac{2}{3}.x=\dfrac{7}{10}-\dfrac{1}{5}=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}:\dfrac{2}{3}=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)

\(c,\left(3\dfrac{4}{5}-2.x\right).1\dfrac{1}{3}=5\dfrac{5}{7}\)

\(\Rightarrow\left(\dfrac{19}{5}-2x\right)\dfrac{4}{3}=\dfrac{40}{7}\)

\(\Rightarrow\left(\dfrac{19}{5}-2x\right)=\dfrac{40}{7}:\dfrac{4}{3}=\dfrac{40}{7}.\dfrac{3}{4}=\dfrac{30}{7}\)

\(\Rightarrow2x=\dfrac{19}{5}-\dfrac{30}{7}=\dfrac{-17}{35}\)

\(\Rightarrow x=\dfrac{-17}{35}:2=\dfrac{-17}{70}\)

\(d,\dfrac{x}{7}=\dfrac{6}{-21}\)

\(\Rightarrow x.\left(-21\right)=6.7\)

\(\Rightarrow x.\left(-21\right)=42\)

\(\Rightarrow x=42:\left(-21\right)=-2\)

a) \(\dfrac{2}{3}\).x + \(\dfrac{1}{2}\)= \(\dfrac{1}{10}\)

=> \(\dfrac{2}{3}\).x = \(\dfrac{1}{10}\)- \(\dfrac{1}{2}\)= \(\dfrac{-2}{5}\)

=> x= \(\dfrac{-2}{5}\): \(\dfrac{2}{3}\)= \(\dfrac{-2}{5}\). \(\dfrac{3}{2}\)

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

Vậy x = \(\dfrac{-3}{5}\)

b) \(\dfrac{2}{3}\).x + \(\dfrac{1}{5}\)= \(\dfrac{7}{10}\)

=> \(\dfrac{2}{3}\).x = \(\dfrac{7}{10}\)- \(\dfrac{1}{5}\)= \(\dfrac{1}{2}\)

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

=> x = \(\dfrac{3}{4}\)

Vậy x = \(\dfrac{3}{4}\)

c) ( \(3\dfrac{4}{5}\)- 2.x) . \(1\dfrac{1}{3}\)= \(5\dfrac{5}{7}\)

=> ( \(\dfrac{19}{5}\) - 2.x) . \(\dfrac{4}{3}\)= \(\dfrac{40}{7}\)

=> ( \(\dfrac{19}{5}\)- 2.x) = \(\dfrac{40}{7}\): \(\dfrac{4}{3}\)= \(\dfrac{40}{7}\). \(\dfrac{3}{4}\)

=> ( \(\dfrac{19}{5}\)- 2.x) = \(\dfrac{30}{7}\)

=> 2.x = \(\dfrac{30}{7}\)- \(\dfrac{19}{5}\)= \(\dfrac{17}{35}\)

=> x = \(\dfrac{17}{35}\) : 2 = \(\dfrac{17}{35}\). \(\dfrac{1}{2}\)

=> x = \(\dfrac{17}{70}\)

Vậy x = \(\dfrac{17}{70}\)

d) \(\dfrac{x}{7}\)= \(\dfrac{6}{-21}\)

=> \(x\)= \(\dfrac{6.7}{-21}\)

=> \(x\)= -2

Mk sửa câu c nha vì mk lm có vấn đề r

c) (\(3\dfrac{4}{5}\)- 2.x) . \(1\dfrac{1}{3}\)=\(5\dfrac{5}{7}\)

=> ( \(\dfrac{19}{5}\)- 2.x) . \(\dfrac{4}{3}\)= \(\dfrac{40}{7}\)

=> ( \(\dfrac{19}{5}\)- 2.x) = \(\dfrac{40}{7}\): \(\dfrac{4}{3}\)= \(\dfrac{40}{7}\). \(\dfrac{3}{4}\)= \(\dfrac{30}{7}\)

=> 2.x = \(\dfrac{19}{5}\)- \(\dfrac{30}{7}\)= \(\dfrac{-17}{35}\)

=> x = \(\dfrac{-17}{35}\): 2 = \(\dfrac{-17}{35}\). \(\dfrac{1}{2}\)= \(\dfrac{-17}{70}\)

Vậy x = \(\dfrac{-17}{70}\)

a) \(\dfrac{2}{3}\cdot x+\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}\cdot x=\dfrac{1}{10}-\dfrac{1}{2}\)

\(\dfrac{2}{3}\cdot x=\dfrac{1}{10}-\dfrac{5}{10}\)

\(\dfrac{2}{3}\cdot x=\dfrac{-2}{5}\)

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

\(x=\dfrac{-2}{5}\cdot\dfrac{3}{2}\)

\(x=\dfrac{-3}{5}\)

Vậy \(x=\dfrac{-3}{5}\).

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

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

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

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

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

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

\(x=\dfrac{3}{4}\)

Vậy \(x=\dfrac{3}{4}\).

c) \(\left(3\dfrac{4}{5}-2\cdot x\right)\cdot1\dfrac{1}{3}=5\dfrac{5}{7}\)

\(\left(\dfrac{19}{5}-2\cdot x\right)\cdot\dfrac{4}{3}=\dfrac{40}{7}\)

\(\dfrac{19}{5}-2\cdot x=\dfrac{40}{7}:\dfrac{4}{3}\)

\(\dfrac{19}{5}-2\cdot x=\dfrac{40}{7}\cdot\dfrac{3}{4}\)

\(\dfrac{19}{5}-2\cdot x=\dfrac{30}{7}\)

\(2\cdot x=\dfrac{19}{5}-\dfrac{30}{7}\)

\(2\cdot x=\dfrac{133}{35}-\dfrac{150}{35}\)

\(2\cdot x=\dfrac{-17}{35}\)

\(x=\dfrac{-17}{35}:2\)

\(x=\dfrac{-17}{35}\cdot\dfrac{1}{2}\)

\(x=\dfrac{-17}{70}\)

Vậy \(x=\dfrac{-17}{70}\).

d) \(\dfrac{x}{7}=\dfrac{6}{-21}\)

\(\dfrac{x}{7}=\dfrac{-6}{21}\)

\(x=\dfrac{-6\cdot7}{21}\)

\(x=-2\)

Vậy \(x=-2\).