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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) \(\left( {{1 \over 2}.\sqrt {{1 \over 2}} – {3 \over 2}.\sqrt 2 + {4 \over 5}.\sqrt {200} } \right):{1 \over 8}\)
d) \(2\sqrt {{{\left( {\sqrt 2 – 3} \right)}^2}} + \sqrt {2.{{\left( { – 3} \right)}^2}} – 5\sqrt {{{\left( { – 1} \right)}^4}} \)
Hướng dẫn làm bài:
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)
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\(\eqalign{
& 0,2\sqrt {{{\left( { – 10} \right)}^2}.3} + 2\sqrt {{{\left( {\sqrt 3 – \sqrt 5 } \right)}^2}} \cr
& = 0,2\left| { – 10} \right|\sqrt 3 + 2\left| {\sqrt 3 – \sqrt 5 } \right| \cr
& = 0,2.10.\sqrt 3 + 2\left( {\sqrt 5 – \sqrt 3 } \right) \cr
& = 2\sqrt 3 + 2\sqrt 5 – 2\sqrt 3 = 2\sqrt 5 \cr} \)
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)
\(\eqalign{
& 2\sqrt {{{\left( {\sqrt 2 – 3} \right)}^2}} + \sqrt {2.{{\left( { – 3} \right)}^2}} – 5\sqrt {{{\left( { – 1} \right)}^4}} \cr
& = 2\left| {\sqrt 2 – 3} \right| + \left| { – 3} \right|\sqrt 2 – 5\left| { – 1} \right| \cr
& = 2\left( {3 – \sqrt 2 } \right) + 3\sqrt 2 – 5 \cr
& = 6 – 2\sqrt 2 + 3\sqrt 2 – 5 = 1 + \sqrt 2 \cr} \)