Đáp án: a.$A=1$
b.$ B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$
Giải thích các bước giải:
a.Ta có:
$A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-6\sqrt{20}}}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-6\sqrt{20}+9}}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}$
$\to A=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}$
$\to A=\sqrt{\sqrt{5}-(\sqrt{5}-1)}$
$\to A=\sqrt{1}$
$\to A=1$
b.Ta có:
$B=\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{3}{\sqrt{x}+1}-\dfrac{2}{x-1}$
$\to B=\dfrac{\sqrt{x}(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{3(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\dfrac{2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{\sqrt{x}(\sqrt{x}+1)-3(\sqrt{x}-1)-2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{x+\sqrt{x}-3\sqrt{x}+3-2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{x-2\sqrt{x}+1}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}-1)(\sqrt{x}+1)}$
$\to B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}$
