giúp tớ làm câu này ạ

Bài 2)

1.

a) $\displaystyle \frac{1-3x}{1+3x} -\frac{1+3x}{1-3x} =\frac{12}{1-9x^{2}} \ ( 1)$

ĐK: $\displaystyle \begin{cases}
1+3x\neq 0 & \\
1-3x\neq 0 & \\
1-9x^{2} \neq 0 & 
\end{cases} \Leftrightarrow x\neq \pm \frac{1}{3}$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
( 1) \Leftrightarrow \frac{( 1-3x)^{2} -( 1+3x)^{2}}{( 1+3x)( 1-3x)} -\frac{12}{1-9x^{2}} =0\\
\Leftrightarrow \frac{( 1-3x-1-3x)( 1-3x+1+3x)}{1-9x^{2}} -\frac{12}{1-9x^{2}} =0\\
\Leftrightarrow \frac{-6x.2-12}{1-9x^{2}} =0\\
\Leftrightarrow \frac{-12x-12}{1-9x^{2}} =0\\
\Leftrightarrow -12x-12=0\\
\Leftrightarrow 12x=-12\\
\Leftrightarrow x=-1
\end{array}$

b) $\displaystyle 5x-\frac{7}{2}( 2x-5) < \frac{2}{3}( x-1)$

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow 5x-7x+\frac{35}{2} < \frac{2}{3} x-\frac{2}{3}\\
\Leftrightarrow -2x-\frac{2}{3} x< \frac{-2}{3} -\frac{35}{2}\\
\Leftrightarrow -\frac{8}{3} x< \frac{-109}{6}\\
\Leftrightarrow x >\frac{109}{16}
\end{array}$