Giải bài 36 trang 66

Đề bài

a) Cho biểu thức $A=\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}$

Chứng minh rằng $A=5$.

b) Cho biểu thức $B=\frac{1}{\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2-\sqrt{3}}}$. Chứng minh rằng $B=\sqrt{6}$.

Lời giải chi tiết

a) $A=\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}$$\begin{array}{c}=\frac{3+\sqrt{8}}{\left(3-\sqrt{8}\right)\left(3-\sqrt{8}\right)}-\frac{\sqrt{8}+\sqrt{7}}{\left(\sqrt{8}-\sqrt{7}\right)\left(\sqrt{8}+\sqrt{7}\right)}+\frac{\sqrt{7}+\sqrt{6}}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}\\ -\frac{\sqrt{6}+\sqrt{5}}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}+\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\\ =\frac{3+\sqrt{8}}{9-8}-\frac{\sqrt{8}+\sqrt{7}}{8-7}+\frac{\sqrt{7}+\sqrt{6}}{7-6}-\frac{\sqrt{6}+\sqrt{5}}{6-5}+\frac{\sqrt{5}+2}{5-4}\\ =3+\sqrt{8}-\sqrt{8}-\sqrt{7}+\sqrt{7}+\sqrt{6}-\sqrt{6}-\sqrt{5}+\sqrt{5}+2\\ =5\end{array}$

Vậy $A=5$.

b) $B=\frac{1}{\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2-\sqrt{3}}}$

$\begin{array}{c}=\frac{\sqrt{2}}{\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}}{\sqrt{4-2\sqrt{3}}}\\ =\frac{\sqrt{2}}{\sqrt{{\left(1+\sqrt{3}\right)}^2}}+\frac{\sqrt{2}}{\sqrt{{\left(1-\sqrt{3}\right)}^2}}\\ =\frac{\sqrt{2}}{\left|1+\sqrt{3}\right|}+\frac{\sqrt{2}}{\left|1-\sqrt{3}\right|}\\ =\frac{\sqrt{2}}{1+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{3}-1}\\ =\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{\left(1+\sqrt{3}\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{\left(\sqrt{3}-1\right)\left(1+\sqrt{3}\right)}\\ =\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{3-1}+\frac{\sqrt{2}\left(1+\sqrt{3}\right)}{3-1}\\ =\frac{\sqrt{6}-\sqrt{2}+\sqrt{2}+\sqrt{6}}{2}\\ =\sqrt{6}\end{array}$

Vậy $B=\sqrt{6}$.