\(\left(\frac{3a}{a^2-4}+\frac{1}{2-a}-\frac{2}{a+2}\right):\left(1-\frac{a^2+4}{a^2-4}\right)\)điều kiện : a khác {-2,2}
=\(\left(\frac{3a}{a^2-4}-\frac{a+2}{a^2-4}-\frac{2a-4}{a^2-4}\right):\left(-\frac{8}{a^2-4}\right)\)
=\(\left(\frac{3a-a-2-2a+4}{a^2-4}\right).\left(\frac{a^2-4}{-8}\right)\)
=\(-\frac{1}{4}\)
\(=\left[\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{\left(a-2\right)}-\frac{2}{\left(a+2\right)}\right]:\left(\frac{a^2-4-a^2-4}{a^2-4}\right)=\left(\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}\right).\frac{\left(a-2\right)\left(a+2\right)}{-8}=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{-8}\)
\(=\frac{-1}{4}\)
\(\left(\frac{3a}{a^2-4}+\frac{1}{2-a}-\frac{2}{a+2}\right):\left(1-\frac{a^2+4}{a^2-4}\right)\)
\(=\left(\frac{3a}{\left(a-2\right)\left(a+2\right)}-\frac{1}{a-2}-\frac{2}{a+2}\right):\left(\frac{a^2-4}{a^2-4}-\frac{a^2+4}{a^2-4}\right)\)
\(=\frac{3a-a-2-2a+4}{\left(a-2\right)\left(a+2\right)}:\frac{\left(-8\right)}{a^2-4}\)
\(=\frac{2}{\left(a-2\right)\left(a+2\right)}.\frac{\left(a-2\right)\left(a+2\right)}{\left(-8\right)}\)
\(=-\frac{1}{4}\)
Quên! còn ĐKXĐ nữa.
ĐKXĐ: a - 2 \(\ne\) 0 và a + 2 \(\ne\) 0
\(\Leftrightarrow\) a\(\ne\) 2 và a \(\ne\) -2
