Đáp án:
Câu 2
$B=\frac{a-3}{2a}$
Câu 3
$C=\frac{1}{2(1+x)}$
Giải thích các bước giải:
Câu 2
$B=\frac{(a+3)^2}{2a^2+6a}.(1-\frac{6a-18}{a^2-9})\\
=\frac{(a+3)^2}{2a(a+3)}.(1-\frac{6(a-3)}{(a-3)(a+3)})\\
DK: {\left\{\begin{aligned}2a\neq 0\\a+3\neq 0\\ a-3 \neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}a \neq 0\\ a \neq 3\\ a \neq -3 \end{aligned}\right.}\\
\Rightarrow \frac{(a+3)^2}{2a(a+3)}.(1-\frac{6}{(a+3)})\\
=\frac{a+3}{2a}.(\frac{a+3}{a+3}-\frac{6}{a+3})\\
=\frac{a+3}{2a}.(\frac{a+3-6}{a+3})\\
=\frac{a+3}{2a}.(\frac{a-3}{a+3})\\
=\frac{(a+3)(a-3)}{2a(a+3)}\\
=\frac{a-3}{2a}$
Câu 3:
$\frac{x}{2x-2}+\frac{x^2+1}{2-2x^2}\\
=\frac{x}{2(x-1)}+\frac{x^2+1}{2(1-x^2)}\\
=\frac{x}{2(x-1)}+\frac{x^2+1}{2(1-x)(1+x)}\\
DK: {\left\{\begin{aligned}x-1\neq 0\\x+1\neq 0\end{aligned}\right.}\Leftrightarrow {\left\{\begin{aligned}x\neq 1\\ x\neq -1\end{aligned}\right.}\\
\Rightarrow C=\frac{x(x+1)}{-2(1-x)(1+x)}+\frac{x^2+1}{2(1-x)(1+x)}\\
=\frac{-x(x+1)+x^2+1}{2(1-x)(1+x)}\\
=\frac{-x^2-x+x^2+1}{2(1-x)(1+x)}\\
=\frac{-x+1}{2(1-x)(1+x)}\\
=\frac{1-x}{2(1-x)(1+x)}\\
=\frac{1}{2(1+x)}$
