Bài tập Thực hiện phép tính lớp 9 (có lời giải)

a) Ta có: \[\sqrt {0,16.81}  = \sqrt {0,16} .\sqrt {81}  = 0,4.9 = 3,6\].

b) Ta có: \(\sqrt {{3^4}.{{\left( { - 5} \right)}^2}}  = \sqrt {{3^4}} .\sqrt {{{\left( { - 5} \right)}^2}}  = {3^2}.\left| { - 5} \right| = 9.5 = 45\).

c)  Ta có: \(\sqrt {16,9.250}  = \sqrt {169.25}  = \sqrt {169} .\sqrt {25}  = 13.5 = 65\).

d) Ta có: \(\sqrt {{5^2}{{.4}^4}}  = \sqrt {{5^2}} .\sqrt {{4^4}}  = {5.4^2} = 80\).

Giải

a) Ta có: \(\sqrt 5 .\sqrt {80}  = \sqrt {5.80}  = \sqrt {400}  = 20\).

b) Ta có: \(\sqrt {2,45} .\sqrt {40} .\sqrt {50}  = \sqrt {2,45.40.50}  = \sqrt {4900}  = 70\).

c)   Ta có: \(\sqrt {0,6} .\sqrt {5,4}  = \sqrt {0,6.5,4}  = \sqrt {3,24}  = 1,8\).

d) Ta có: \(\sqrt {8,1} .\sqrt 5 .\sqrt {4,5}  = \sqrt {8,1.5.4,5}  = \sqrt {182,25}  = 13,5\).

Ta có: \(\sqrt {13.25.52}  = \sqrt {16\,\,900}  = 130\) (chọn B).

a) Ta có: \(\sqrt {{{25}^2} - {{24}^2}}  = \sqrt {\left( {25 - 24} \right)\left( {25 + 24} \right)}  = \sqrt {49}  = 7\).

b)  Ta có: \[\sqrt {{{26}^2} - {{10}^2}}  = \sqrt {\left( {26 - 10} \right)\left( {26 + 10} \right)}  = \sqrt {16.36}  = 4.6 = 24\].

c)   Ta có: \(\sqrt {{{137}^2} - {{88}^2}}  = \sqrt {\left( {137 - 88} \right)\left( {137 + 88} \right)}  = \sqrt {49.225}  = 7.15 = 105\).

d)  Ta có: \(\sqrt {{{481}^2} - {{480}^2}}  = \sqrt {\left( {481 - 480} \right)\left( {481 + 480} \right)}  = \sqrt {961}  = 31\).

a) Ta có: \[\sqrt {\frac{{289}}{{225}}}  = \frac{{\sqrt {289} }}{{\sqrt {225} }} = \frac{{17}}{{15}}\].

b) Ta có \[\sqrt {2\frac{{14}}{{25}}}  = \sqrt {\frac{{64}}{{25}}}  = \frac{{\sqrt {64} }}{{\sqrt {25} }} = \frac{8}{5}\].

c)  Ta có: \[\sqrt {\frac{{0,25}}{9}}  = \frac{{\sqrt {0,25} }}{{\sqrt 9 }} = \frac{{0,5}}{9}\].

d) \[\sqrt {\frac{{8,1}}{{1,6}}}  = \sqrt {\frac{{81}}{{16}}}  = \frac{{\sqrt {81} }}{{\sqrt {16} }} = \frac{9}{4}\].

a) Ta có: \[\frac{{\sqrt 2 }}{{\sqrt {18} }} = \sqrt {\frac{2}{{18}}}  = \sqrt {\frac{1}{9}}  = \frac{1}{3}\].                                       Ta có \[\frac{{\sqrt {15} }}{{\sqrt {735} }} = \sqrt {\frac{{15}}{{735}}}  = \sqrt {\frac{1}{{49}}}  = \frac{1}{7}\].

b) Ta có: \[\frac{{\sqrt {12500} }}{{\sqrt {500} }} = \sqrt {\frac{{12500}}{{500}}}  = \sqrt {25}  = 5\].                                                            \[\sqrt {\frac{{{6^5}}}{{{2^3}{{.3}^5}}}}  = \sqrt {\frac{{{2^5}{{.3}^5}}}{{{2^3}{{.3}^5}}}}  = \sqrt 4  = 2\].

a) Ta có: \[\sqrt {1\frac{9}{{16}}.5\frac{4}{5}.0,01}  = \sqrt {\frac{{25}}{{16}}} .\sqrt {\frac{{49}}{9}} .\sqrt {0,01}  = \frac{5}{4}.\frac{7}{3}.0,1 = \frac{7}{{24}}\].

b) Ta có \[\sqrt {1,44.1,21 - 1,44.0,4}  = \sqrt {1,44.\left( {1,21 - 0,4} \right)} \]

\[ = \sqrt {1,44.0,81}  = \sqrt {1,44} .\sqrt {0,81}  = 1,2.0,9 = 1,08\].

c)  Ta có: \[\sqrt {\frac{{{{165}^2} - {{124}^2}}}{{164}}}  = \sqrt {\frac{{\left( {165 - 124} \right)\left( {165 + 124} \right)}}{{164}}}  = \sqrt {\frac{{41.289}}{{164}}}  = \sqrt {\frac{{289}}{4}}  = \frac{{17}}{2}\].

d) \[\sqrt {\frac{{{{149}^2} - {{76}^2}}}{{{{457}^2} - {{384}^2}}}}  = \sqrt {\frac{{73.225}}{{73.841}}}  = \frac{{\sqrt {225} }}{{\sqrt {841} }} = \frac{{15}}{{29}}\].