Ta có :
\(\left(x-5\right)\left(6x+1\right)-\left(2x-3\right)\left(3x+4\right)-x=6\)
\(\Rightarrow\left(6x^2-30x+x-5\right)-\left(6x^2-9x+8x-12\right)-x=6\)
\(\Rightarrow6x^2-29x-5-\left(6x^2-x-12\right)-x=6\)
\(\Rightarrow6x^2-29x-5-6x^2+x+12-x=6\)
\(\Rightarrow\left(6x^2-6x^2\right)+\left(-29x+x-x\right)+\left(12-5\right)=6\)
\(\Rightarrow-29x+7=6\)
\(\Rightarrow-29x=6-7\)
\(\Rightarrow-29x=-1\)
\(\Rightarrow x=\frac{1}{29}\)
Vậy \(x=\frac{1}{29}\)
Tìm x :
\(\left(x-5\right)\cdot\left(6x+1\right)-\left(2x-3\right)\cdot\left(3x+4\right)-x=6\)
\(=\left(x-5\right)\cdot\left(4x-2\right)\cdot\left(2x+4\right)=6\)
\(=\left(x-5\right)\cdot\left(4\cdot x-2\right)\cdot\left(2\cdot x+4\right)=6\)
\(=\left(x-5-2+4\right)\cdot4x\cdot2x=6\)
\(=\left(x-3\right)\cdot8x=6\)
\(9x=6+3=9\)
\(x=9:9=1\)
