\(\dfrac{3\left(x-11\right)}{4}=\dfrac{3\left(x+1\right)}{5}\\ \Rightarrow\dfrac{3x-33}{4}=\dfrac{3x+3}{5}=\dfrac{3x-33-3x-3}{4-5}\\ =\dfrac{-36}{-1}=36\\ \Rightarrow3\left(x-11\right)=144\\ \Rightarrow x-11=48\\ \Rightarrow x=59\)
Đề như vậy thì chỉ cần 2 pth đầu là đc chắc bn ghi đề sai, mk lm theo đề như này nhé:
\(\dfrac{3\left(x-11\right)}{4}+\dfrac{3\left(x+1\right)}{5}=\dfrac{2\left(2x-5\right)}{10}\)
\(\Leftrightarrow\dfrac{15\left(x-11\right)}{20}+\dfrac{12\left(x+1\right)}{20}=\dfrac{4\left(2x-5\right)}{20}\)
\(\Leftrightarrow15\left(x-11\right)+12\left(x+1\right)=4\left(2x-5\right)\)
\(\Leftrightarrow15x-165+12x+12=8x-20\)
\(\Leftrightarrow15x+12x-8x=-20-12+165\)
\(\Leftrightarrow19x=133\Leftrightarrow x=7\)
Vậy x = 7
