a) Ta có: \(\frac{2\left(x-4\right)}{3}+\frac{3x+13}{8}=\frac{2\left(2x-3\right)}{5}+12\)
⇔\(\frac{80\left(x-4\right)}{120}+\frac{15\left(3x+13\right)}{120}-\frac{48\left(2x-3\right)}{120}-\frac{1440}{120}=0\)
⇔\(80\left(x-4\right)+15\left(3x+13\right)-48\left(2x-3\right)-1440=0\)
\(\Leftrightarrow80x-320+45x+195-96x+144-1440=0\)
⇔\(29x-1421=0\)
\(\Leftrightarrow29x=1421\)
hay x=49
Vậy: x=49
b) Ta có: \(\frac{2\left(5x+2\right)}{9}-1=\frac{4\left(33+2x\right)}{5}-\frac{5\left(1-11x\right)}{9}\)
⇔\(\frac{10\left(5x+2\right)}{45}-\frac{45}{45}-\frac{36\left(33+2x\right)}{45}+\frac{25\left(1-11x\right)}{45}=0\)
⇔\(10\left(5x+2\right)-45-36\left(33+2x\right)+25\left(1-11x\right)=0\)
\(\Leftrightarrow50x+20-45-1188-72x+25-275x=0\)
\(\Leftrightarrow-297x-1188=0\)
\(\Leftrightarrow-297x=1188\)
hay x=-4
Vậy: x=-4
