# Bài 70 trang 40 SGK Toán 9 tập 1

##### Hướng dẫn giải

$$\begin{array}{l} \sqrt {AB} = \sqrt A .\sqrt B \,\,\left( {A \ge 0,B \ge 0} \right)\\ \sqrt {{A^2}} = \left| A \right| \end{array}$$

Lời giải chi tiết

a)

\eqalign{ & \sqrt {{{25} \over {81}}.{{16} \over {49}}.{{196} \over 9}} \cr & = \sqrt {{{25} \over {81}}} .\sqrt {{{16} \over {49}}} .\sqrt {{{196} \over 9}} \cr & = {5 \over 9}.{4 \over 7}.{{14} \over 3} = {{40} \over {27}} \cr}

b)

\eqalign{ & \sqrt {3{1 \over {16}}.2{{14} \over {25}}2{{34} \over {81}}} \cr & = \sqrt {{{49} \over {16}}.{{64} \over {25}}.{{196} \over {81}}} \cr & = \sqrt {{{49} \over {16}}} .\sqrt {{{64} \over {25}}} .\sqrt {{{196} \over {81}}} \cr & = {7 \over 4}.{8 \over 5}.{{14} \over 9} = {{196} \over {45}} \cr}

c)

$$\begin{array}{l} \frac{{\sqrt {640} .\sqrt {34,3} }}{{\sqrt {567} }} = \sqrt {\frac{{640.34,3}}{{567}}} = \sqrt {\frac{{64.49.7}}{{81.7}}} \\ = \sqrt {\frac{{64.49}}{{81}}} = \frac{{\sqrt {64} .\sqrt {49} }}{{\sqrt {81} }} = \frac{{8.7}}{9} = \frac{{56}}{9} \end{array}$$

d)

\eqalign{ & \sqrt {21,6} .\sqrt {810.} \sqrt {{{11}^2} - {5^2}} \cr & = \sqrt {21,6.810.\left( {{{11}^2} - {5^2}} \right)} \cr & = \sqrt {216.81.\left( {11 + 5} \right)\left( {11 - 5} \right)} \cr & = \sqrt {{36.6}{{.9}^2}{{.4}^2}.6} = 36.9.4 = 1296 \cr}