\(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(\Rightarrow4x-3-x-5=30-3x\)
\(\Rightarrow3x-8=30-3x\)
\(\Rightarrow6x=38\)
\(\Rightarrow x=\dfrac{19}{3}\)
Vậy \(x=\dfrac{19}{3}\)
\(Ta \) \(có:\)
\(\left(4x-3\right)-\left(x-5\right)=3\left(10-x\right)\)
\(\Leftrightarrow4x-3-x+5\)
\(\Leftrightarrow\left(4x-x\right)+\left(-3+5\right)=30-3x\)
\(\Leftrightarrow3x+2=30-x\)
\(\Leftrightarrow\left(3x+2\right)+x=30\)
\(\Leftrightarrow3x+2+x=30\)
\(\Leftrightarrow\left(3x+x\right)+2=30\)
\(\Leftrightarrow4x+2=30\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(Vậy \) \(x=7\\\)
