Bài 1:
a) Ta có: 22x-13=x-6
\(\Leftrightarrow22x-13-x+6=0\)
\(\Leftrightarrow21x-7=0\)
\(\Leftrightarrow21x=7\)
hay \(x=\frac{1}{3}\)
Vậy: \(x=\frac{1}{3}\)
b) Ta có: (x-7)(2x+10)=0
\(\Leftrightarrow\left(x-7\right)\cdot2\cdot\left(x+5\right)=0\)
mà \(2\ne0\)
nên \(\left[{}\begin{matrix}x-7=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{-5;7\right\}\)
c) ĐKXĐ: \(x\ne14\)
Ta có: \(\frac{12x+9}{x-14}=7\)
\(\Leftrightarrow12x+9=7\left(x-14\right)\)
\(\Leftrightarrow12x+9=7x-98\)
\(\Leftrightarrow12x+9-7x+98=0\)
\(\Leftrightarrow5x+107=0\)
\(\Leftrightarrow5x=-107\)
hay \(x=\frac{-107}{5}\)(tm)
Vậy: \(x=\frac{-107}{5}\)
d) Ta có: \(\frac{x+2}{4}+\frac{3x-4}{6}=\frac{x-14}{24}\)
\(\Leftrightarrow\frac{6\left(x+2\right)}{24}+\frac{4\left(3x-4\right)}{24}=\frac{x-14}{24}\)
Suy ra: \(6\left(x+2\right)+4\left(3x-4\right)=x-14\)
\(\Leftrightarrow6x+12+12x-16-x+14=0\)
\(\Leftrightarrow17x+10=0\)
\(\Leftrightarrow17x=-10\)
hay \(x=\frac{-10}{17}\)
Vậy: \(x=\frac{-10}{17}\)
