Question

G=\(\left(\frac{a\sqrt{a}-3}{a-2\sqrt{a}-3}-\frac{2\sqrt{a-3}}{\sqrt{a}+1}+\frac{\sqrt{a}+3}{3-\sqrt{a}}\right)\left(\frac{a+8}{a-1}\right)\)

Answer 1

Lời giải:
ĐKXĐ: a\geq 0; a\neq 1; a\neq 9$

Ta có:

\(G=\left[\frac{a\sqrt{a}-3}{(\sqrt{a}+1)(\sqrt{a}-3)}-\frac{2(\sqrt{a}-3)^2}{(\sqrt{a}+1)(\sqrt{a}-3)}-\frac{(\sqrt{a}+3)(\sqrt{a}+1)}{(\sqrt{a}-3)(\sqrt{a}+1)}\right].\frac{a+8}{a-1}\)

\(=\frac{a\sqrt{a}-3-2(\sqrt{a}-3)^2-(\sqrt{a}+3)(\sqrt{a}+1)}{(\sqrt{a}+1)(\sqrt{a}-3)}.\frac{a+8}{a-1}\)

\(=\frac{a\sqrt{a}-3a+8\sqrt{a}-24}{(\sqrt{a}+1)(\sqrt{a}-3)}.\frac{a+8}{a-1}=\frac{(\sqrt{a}-3)(a+8)}{(\sqrt{a}+1)(\sqrt{a}-3)}.\frac{a+8}{a-1}=\frac{(a+8)^2}{(\sqrt{a}+1)(a-1)}\)