Giải bài 47 trang 68

Đề bài

Rút gọn biểu thức

a) $\left(5\sqrt{\frac{1}{5}}-\frac{1}{2}\sqrt{20}+\sqrt{5}\right)\sqrt{5}$

b) $\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{9}{7}}+\sqrt{7}\right):\sqrt{7}$

c) ${\left(\sqrt{\frac{2}{3}}-\sqrt{\frac{3}{2}}\right)}^2$

d) $\frac{\sqrt{{312}^2-{191}^2}}{\sqrt{503}}$

e) $\sqrt{27.{\left(1-\sqrt{3}\right)}^4}:3\sqrt{15}$

g) $\frac{\sqrt[3]{3135}}{\sqrt[3]{35}}-\sqrt[3]{354}.\sqrt[3]{34}$

Lời giải chi tiết

a) $\left(5\sqrt{\frac{1}{5}}-\frac{1}{2}\sqrt{20}+\sqrt{5}\right)\sqrt{5}$

$=\left(5\frac{1}{\sqrt{5}}-\frac{1}{2}.2.\sqrt{5}+\sqrt{5}\right)\sqrt{5}=5-5+5=5.$

b) $\left(\sqrt{\frac{1}{7}}-\sqrt{\frac{9}{7}}+\sqrt{7}\right):\sqrt{7}$

$=\left(\frac{1}{\sqrt{7}}-\frac{3}{\sqrt{7}}+\sqrt{7}\right).\frac{1}{\sqrt{7}}=\frac{1}{7}-\frac{3}{7}+1=\frac{5}{7}.$

c) ${\left(\sqrt{\frac{2}{3}}-\sqrt{\frac{3}{2}}\right)}^2$

$=\frac{2}{3}-2\sqrt{\frac{2}{3}.\frac{3}{2}}+\frac{3}{2}=\frac{13}{6}-2=\frac{1}{6}$

d) $\frac{\sqrt{{312}^2-{191}^2}}{\sqrt{503}}$

$=\frac{\sqrt{\left(312-191\right)\left(312+191\right)}}{\sqrt{503}}$

$=\frac{\sqrt{121.503}}{\sqrt{503}}=\sqrt{121}=11$

e) $\sqrt{27.{\left(1-\sqrt{3}\right)}^4}:3\sqrt{15}$

$=3.\sqrt{3.}{\left(1-\sqrt{3}\right)}^2.\frac{1}{3\sqrt{15}}=\frac{{\left(1-\sqrt{3}\right)}^2}{\sqrt{5}}$

$=\frac{\sqrt{5}\left(1-2\sqrt{3}+3\right)}{5}=\frac{\sqrt{5}\left(4-2\sqrt{3}\right)}{5}=\frac{4\sqrt{5}-2\sqrt{15}}{5}$

g) $\frac{\sqrt[3]{3135}}{\sqrt[3]{35}}-\sqrt[3]{354}.\sqrt[3]{34}$

$=\frac{\sqrt[3]{35}.\sqrt[3]{327}}{\sqrt[3]{35}}-\sqrt[3]{32.27}.\sqrt[3]{34}$

$=3-3\sqrt[3]{32}.\sqrt[3]{34}=3-3\sqrt[3]{38}=3-3.2=-3$