Question

tìm x biết a, \(\dfrac{2}{3}\)-\(\dfrac{5}{12}\)x=\(\dfrac{-8}{3}\)

b, 3x-2.(2x-1)=\(1\dfrac{1}{3}\)

c,\(\dfrac{x+4}{20}\)=\(\dfrac{5}{x+4}\)

Làm ơn giúp mk vs nhamk đg cần nhưng còn nhìu bài để làm cho nên giúp mk vs nha Thanks trước

Answer 1

a)

\(\dfrac{2}{3}-\dfrac{5}{12}x=\dfrac{-8}{3}\)\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}-\left(-\dfrac{8}{3}\right)\)

\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\)

\(\Rightarrow x=\dfrac{10}{3}:\dfrac{5}{12}=8\)

b) \(3x-2\left(2x-1\right)=1\dfrac{1}{3}\)\(\Rightarrow3x-4x+2=\dfrac{4}{3}\)

\(\Rightarrow3x-4x=\dfrac{4}{3}-2\)

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

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\Rightarrow\left(x+4\right)\left(x+4\right)=20.5\)

\(\Rightarrow\left(x+4\right)^2=100\)

\(\Rightarrow\left(x+4\right)^2=10^2\) hoặc \(\left(x+4\right)^2=\left(-10\right)^2\)

=> x+4=10 => x+4=-10

=> x=6 => x=-14