Lúc nãy mình để 2 tab bài gần nhau nên post nhầm bài.
Lời giải:
a) Để $A,B$ có nghĩa thì \(\left\{\begin{matrix} a+2\geq 0\\ 1-\sqrt{a+2}\neq 0\\ 1+\sqrt{a+2}\neq 0\\ 1+a^3\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} a\geq -2\\ a\neq -1\\ \end{matrix}\right.\)
b)
\(A=\frac{1-\sqrt{a+2}}{2(1+\sqrt{a+2})(1-\sqrt{a+2})}+\frac{1+\sqrt{a+2}}{2(1-\sqrt{a+2})(1+\sqrt{a+2})}\)
\(=\frac{2}{2(1+\sqrt{a+2})(1-\sqrt{a+2})}=\frac{1}{1-(a+2)}=\frac{-1}{a+1}\)
\(B=A+\frac{a^2-2a}{a^3+1}=\frac{-1}{a+1}+\frac{a^2-2a}{(a+1)(a^2-a+1)}=\frac{-(a^2-a+1)+a^2-2a}{(a+1)(a^2-a+1)}=\frac{-(a+1)}{(a+1)(a^2-a+1)}=\frac{-1}{a^2-a+1}\)
c)
Ta thấy \(a^2-a+1=(a-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}, \forall a\geq -2; a\neq -1\)
\(\Rightarrow \frac{1}{a^2-a+1}\leq \frac{1}{\frac{3}{4}}=\frac{4}{3}\)
\(\Rightarrow B=-\frac{1}{a^2-a+1}\geq \frac{-4}{3}\)
Vậy $B_{\min}=\frac{-4}{3}$ khi $(a-\frac{1}{2})^2=0\Leftrightarrow a=\frac{1}{2}$
