Câu 2: Trang 28 sách phát triển năng lực toán 9 tập 1
Rút gọn các biểu thức:
a, $\frac{1}{3+\sqrt{5}}+\frac{1}{3-\sqrt{5}}$;
b, $\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}$;
c, $\frac{4}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}$;
d, $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$;
e, $\frac{5-\sqrt{5}}{5+\sqrt{5}}+\frac{5+\sqrt{5}}{5-\sqrt{5}}$;
f, $\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}$.
Bài Làm:
a, $\frac{1}{3+\sqrt{5}}+\frac{1}{3-\sqrt{5}}$ = $\frac{1.(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}+\frac{1.(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$
= $\frac{3-\sqrt{5}+3+\sqrt{5}}{9-5}$ = $\frac{6}{4}=\frac{3}{2}$
b, $\frac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}$ = $\frac{2.5+2\sqrt{2.5}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}$
= $\frac{2.\sqrt{5}.\sqrt{5}+2\sqrt{2}.\sqrt{5}}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}$
= $\frac{2\sqrt{5}(\sqrt{5}+\sqrt{2})}{\sqrt{5}+\sqrt{2}}+\frac{8}{1-\sqrt{5}}$
= $2\sqrt{5}+\frac{8}{1-\sqrt{5}}$ = $2\sqrt{5}+\frac{8.(1+\sqrt{5})}{1-5}$ = $2\sqrt{5} - 2.(1+\sqrt{5})$ = -2
c, $\frac{4}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}$ = $\frac{4.(\sqrt{3}-1)}{3-1}+\frac{1.(\sqrt{3}+2)}{3-4}+\frac{6.(\sqrt{3}+3)}{3-9}$
= $2.(\sqrt{3}-1)-(\sqrt{3}+2)-(\sqrt{3}+3)$ = -7
d, $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$ = $\frac{\sqrt{2}.(2+\sqrt{3})}{\sqrt{2}.(\sqrt{2}+\sqrt{2+\sqrt{3}})}+\frac{\sqrt{2}.(2-\sqrt{3})}{\sqrt{2}.(\sqrt{2}-\sqrt{2-\sqrt{3}})}$
= $\frac{\sqrt{2}.(2+\sqrt{3})}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}.(2-\sqrt{3})}{2-\sqrt{4-2\sqrt{3}}}$ = $\frac{\sqrt{2}.(2+\sqrt{3})}{2+\sqrt{3+2\sqrt{3}+1}}+\frac{\sqrt{2}.(2-\sqrt{3})}{2-\sqrt{3-2\sqrt{3}+1}}$
= $\frac{\sqrt{2}.(2+\sqrt{3})}{2+\sqrt{(\sqrt{3}+1)^{2}}}+\frac{\sqrt{2}.(2-\sqrt{3})}{2-\sqrt{(\sqrt{3}-1)^{2}}}$ = $\frac{\sqrt{2}.(2+\sqrt{3})}{2+\sqrt{3}+1}+\frac{\sqrt{2}.(2-\sqrt{3})}{2-\sqrt{3}+1}$
= $\frac{\sqrt{2}.(2+\sqrt{3})}{3+\sqrt{3}}+\frac{\sqrt{2}.(2-\sqrt{3})}{3-\sqrt{3}}$ = $\frac{\sqrt{2}.(2+\sqrt{3}).(3-\sqrt{3})}{9-3}+\frac{\sqrt{2}.(2-\sqrt{3}).(3+\sqrt{3})}{9-3}$
= $\frac{\sqrt{2}.(2+\sqrt{3}).(3-\sqrt{3})+\sqrt{2}.(2-\sqrt{3}).(3+\sqrt{3})}{6}$ = $\frac{6\sqrt{2}}{6}$ = $\sqrt{2}$
e, $\frac{5-\sqrt{5}}{5+\sqrt{5}}+\frac{5+\sqrt{5}}{5-\sqrt{5}}$ = $\frac{(5-\sqrt{5}).(5-\sqrt{5})}{25-5}+\frac{(5+\sqrt{5}).(5+\sqrt{5})}{25-5}$
= $\frac{(25-10\sqrt{5}+5)+(25+10\sqrt{5}+5)}{20}$ = $\frac{60}{20}$ = 3
f, $\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}$ = $\frac{\sqrt{3}.(\sqrt{\sqrt{3}+1}+1)}{\sqrt{3}+1-1}-\frac{\sqrt{3}.(\sqrt{\sqrt{3}+1}-1)}{\sqrt{3}+1-1}$
= $\frac{\sqrt{3}.(\sqrt{\sqrt{3}+1}+1)}{\sqrt{3}}-\frac{\sqrt{3}.(\sqrt{\sqrt{3}+1}-1)}{\sqrt{3}}$ = $\sqrt{\sqrt{3}+1}+1- (\sqrt{\sqrt{3}+1}-1)$ = 2