Ta có \(\frac{-\frac{2}{3}+\frac{3}{4}-2}{\frac{2}{3}-\frac{3}{4}-2}.\frac{-\frac{2}{3}-\frac{3}{4}-2}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\frac{-\left(\frac{2}{3}-\frac{3}{4}+2\right)}{\frac{2}{3}-\frac{3}{4}-2}.\frac{-\left(\frac{2}{3}+\frac{3}{4}+2\right)}{\frac{2}{3}+\frac{3}{4}+2}\)
\(=\frac{23}{25}.\left(-1\right)\)
\(=\frac{-23}{25}\)