7 bài tập Thực hiện phép tính (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: $\sqrt {8{,}1} .\sqrt 5 .\sqrt {4{,}5} = \sqrt {8{,}1.5.4{,}5} = \sqrt {182{,}25} = 13{,}5$.

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