Bài 71 trang 40 SGK Toán 9 tập 1

Rút gọn các biểu thức sau:

a)  \(\left( {\sqrt 8  - 3.\sqrt 2  + \sqrt {10} } \right)\sqrt 2  - \sqrt 5 \) 

b)  \(0,2\sqrt {{{\left( { - 10} \right)}^2}.3}  + 2\sqrt {{{\left( {\sqrt 3  - \sqrt 5 } \right)}^2}} \)

c)  \(\displaystyle \left( {{1 \over 2}.\sqrt {{1 \over 2}}  - {3 \over 2}.\sqrt 2  + {4 \over 5}.\sqrt {200} } \right):{1 \over 8}\)

d)  \(2\sqrt {{{\left( {\sqrt 2  - 3} \right)}^2}}  + \sqrt {2.{{\left( { - 3} \right)}^2}}  - 5\sqrt {{{\left( { - 1} \right)}^4}} \)

a)\(\eqalign{

& \left( {\sqrt 8 - 3.\sqrt 2 + \sqrt {10} } \right)\sqrt 2 - \sqrt 5 \cr
& = \sqrt {16} - 6 + \sqrt {20} - \sqrt 5 \cr
& = 4 - 6 + 2\sqrt 5 - \sqrt 5 = - 2 + \sqrt 5 \cr} \)                         

b) 

\(\eqalign{
& 0,2\sqrt {{{\left( { - 10} \right)}^2}.3} + 2\sqrt {{{\left( {\sqrt 3 - \sqrt 5 } \right)}^2}} \cr
& = 0,2\left| { - 10} \right|\sqrt 3 + 2\left| {\sqrt 3 - \sqrt 5 } \right| \cr
& = 0,2.10.\sqrt 3 + 2\left( {\sqrt 5 - \sqrt 3 } \right) \cr
& = 2\sqrt 3 + 2\sqrt 5 - 2\sqrt 3 = 2\sqrt 5 \cr} \)

(vì \(- 10 < 0;\sqrt 3  < \sqrt 5  \Leftrightarrow \sqrt 3  - \sqrt 5  < 0\))

c)  

\(\eqalign& \left( {{1 \over 2}.\sqrt {{1 \over 2}} - {3 \over 2}.\sqrt 2 + {4 \over 5}.\sqrt {200} } \right):{1 \over 8} \cr & = \left( {{1 \over 2}\sqrt {{2 \over {{2^2}}}} - {3 \over 2}\sqrt 2 + {4 \over 5}\sqrt {{{10}^2}.2} } \right):{1 \over 8} \cr
& = \left( {{1 \over 4}\sqrt 2 - {3 \over 2}\sqrt 2 + 8\sqrt 2 } \right):{1 \over 8} \cr
& = {{27} \over 4}\sqrt 2 .8 = 54\sqrt 2 \cr} \)                  

d)  \(\eqalign{& 2\sqrt {{{\left( {\sqrt 2 - 3} \right)}^2}} + \sqrt {2.{{\left( { - 3} \right)}^2}} - 5\sqrt {{{\left( { - 1} \right)}^4}} \cr 

& = 2\left| {\sqrt 2 - 3} \right| + \left| { - 3} \right|\sqrt 2 - 5.(-1)^2 \cr
& = 2\left( {3 - \sqrt 2 } \right) + 3\sqrt 2 - 5 \cr
& = 6 - 2\sqrt 2 + 3\sqrt 2 - 5 = 1 + \sqrt 2 \cr} \)