Question

\(\left(\dfrac{3}{4}x+5\right)-\left(\dfrac{2}{3}x-4\right)-\left(\dfrac{1}{6}x+1\right)=\left(\dfrac{1}{3}x+4\right)-\left(\dfrac{1}{3}x-3\right)\)

Tìm x

Answer 1

\(\left(\dfrac{3}{4}x+5\right)-\left(\dfrac{2}{3}x-4\right)-\left(\dfrac{1}{6}x+1\right)=\left(\dfrac{1}{3}x+4\right)-\left(\dfrac{1}{3}x-3\right)\)

\(\Rightarrow\dfrac{3}{4}x+5-\dfrac{2}{3}x+4-\dfrac{1}{6}x-1=\dfrac{1}{3}x+4-\dfrac{1}{3}x+3\)

\(\Rightarrow\left(\dfrac{3}{4}x-\dfrac{2}{3}x-\dfrac{1}{6}x\right)+\left(5+4-1\right)=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(4+3\right)\)

\(\Rightarrow\left(\dfrac{9}{12}x-\dfrac{8}{12}x-\dfrac{2}{12}x\right)+8=7\)

\(\Rightarrow-\dfrac{1}{12}x+8=7\)

\(\Rightarrow-\dfrac{1}{12}x=-1\)

\(\Rightarrow x=12\)