Giải bài 47 trang 68 sách bài tập toán 9 - Cánh diều tập 1

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) $\frac{{\sqrt[3]{{135}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{54}}.\sqrt[3]{4}$

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) $\frac{{\sqrt[3]{{135}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{54}}.\sqrt[3]{4}$

$= \frac{{\sqrt[3]{5}.\sqrt[3]{{27}}}}{{\sqrt[3]{5}}} - \sqrt[3]{{2.27}}.\sqrt[3]{4}$

$= 3 - 3\sqrt[3]{2}.\sqrt[3]{4} = 3 - 3\sqrt[3]{8} = 3 - 3.2 =  - 3$