Đáp án:
a) 3x+6
b) \(
\frac{1}{{x - 2}}
\)
Giải thích các bước giải:
\(
\begin{array}{l}
a)\frac{{(x + 2)^2 }}{{x - 1}}:\frac{{x + 2}}{{3x - 3}} \\
Đk:x \ne 1;x \ne - 2 \\
= \frac{{(x + 2)^2 }}{{x - 1}}.\frac{{3x - 3}}{{x + 2}} \\
= \frac{{(x + 2)^2 }}{{x - 1}}.\frac{{3(x - 1)}}{{x + 2}} \\
= 3(x + 2) = 3x + 6 \\
b)\frac{2}{{x - 2}} + \frac{4}{{x + 2}} - \frac{{5x - 6}}{{x^2 - 4}} \\
Đk:x \ne \pm 2 \\
= \frac{{2(x + 2)}}{{(x - 2)(x + 2)}} + \frac{{4(x - 2)}}{{(x + 2)(x - 2)}} - \frac{{5x - 6}}{{(x - 2)(x + 2)}} \\
= \frac{{2x + 4 + 4x - 8 - 5x + 6}}{{(x - 2)(x + 2)}} \\
= \frac{{x + 2}}{{(x - 2)(x + 2)}} = \frac{1}{{x - 2}} \\
\end{array}
\)
