Question

\(\frac{11}{4-\sqrt{5}}+\frac{4}{3-\sqrt{5}}-\frac{19}{\sqrt{21+4\sqrt{5}}}\)

Answer 1

Lời giải:

Đặt biểu thức là $A$.

$21+4\sqrt{5}=20+1+2.\sqrt{20}.\sqrt{1}=(\sqrt{20}+1)^2$

$\Rightarrow \sqrt{21+4\sqrt{5}}=\sqrt{20}+1$

Do đó:

$A=\frac{11}{4-\sqrt{5}}+\frac{4}{3-\sqrt{5}}-\frac{19}{\sqrt{20}+1}$
$=\frac{11(4+\sqrt{5})}{(4-\sqrt{5})(4+\sqrt{5})}+\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{19(\sqrt{20}-1)}{(\sqrt{20}+1)(\sqrt{20}-1)}$

$=\frac{11(4+\sqrt{5})}{11}+\frac{4(3+\sqrt{5})}{4}-\frac{19(\sqrt{20}-1)}{19}$

$=4+\sqrt{5}+3+\sqrt{5}-(\sqrt{20}-1)$

$=8$