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Rút gọn :
a) \(\sqrt {5 + 2\sqrt 6 } – \sqrt {5 – 2\sqrt 6 } \);
b) \(\sqrt {8 + \sqrt {60} } – \sqrt {8 – 2\sqrt {15} } \);
c) \(\sqrt {4 – \sqrt 7 } + \sqrt {4 + \sqrt 7 } \);
d) \(\sqrt {5\sqrt 3 + 5\sqrt {48 – 10\sqrt {7 + 4\sqrt 3 } } } \);
e) \(\left( {x – 4} \right)\sqrt {16 – 8x + {x^2}} \) với \(x \ge 4\);
f) \(\left( {2x – 5} \right)\sqrt {\dfrac{2}{{{{\left( {2x – 5} \right)}^2}}}} \) với \(x \ne \dfrac{5}{2}\);
g) \(\sqrt {x – 4\sqrt {x – 4} } \) với \(x \ge 4\).
+) Sử dụng công thức; \(\sqrt {{A^2}} = \left| A \right| = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ – A\;\;khi\;A < 0\end{array} \right..\)
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\(\begin{array}{l}a)\;\sqrt {5 + 2\sqrt 6 } – \sqrt {5 – 2\sqrt 6 } = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 3 .\sqrt 2 + {{\left( {\sqrt 2 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 } \right)}^2} – 2.\sqrt 3 .\sqrt 2 + {{\left( {\sqrt 2 } \right)}^2}} \\ = \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}} = \left| {\sqrt 3 + \sqrt 2 } \right| – \left| {\sqrt 3 – \sqrt 2 } \right|\\ = \sqrt 3 + \sqrt 2 – \sqrt 3 + \sqrt 2 = 2\sqrt 2 \;\;\;\left( {do\;\;\sqrt 3 > \sqrt 2 } \right).\end{array}\)
\(\begin{array}{l}b)\;\sqrt {8 + \sqrt {60} } – \sqrt {8 – 2\sqrt {15} } = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2.\sqrt 5 .\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2.\sqrt 5 .\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} \\ = \sqrt {{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 5 – \sqrt 3 } \right)}^2}} = \left| {\sqrt 5 + \sqrt 3 } \right| – \left| {\sqrt 5 – \sqrt 3 } \right|\\ = \sqrt 5 + \sqrt 3 – \sqrt 5 + \sqrt 3 = 2\sqrt 3 \;\;\left( {do\;\;\sqrt 5 > \sqrt 3 } \right).\end{array}\)
\(\begin{array}{l}c)\;\sqrt {4 – \sqrt 7 } + \sqrt {4 + \sqrt 7 } = \dfrac{{\sqrt 2 .\sqrt {4 – \sqrt 7 } }}{{\sqrt 2 }} + \dfrac{{\sqrt 2 .\sqrt {4 + \sqrt 7 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {8 – 2\sqrt 7 } }}{{\sqrt 2 }} + \dfrac{{\sqrt {8 + 2\sqrt 7 } }}{{\sqrt 2 }} = \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} – 2\sqrt 7 + 1} }}{{\sqrt 2 }} + \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7 + 1} }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left( {\sqrt 7 – 1} \right)}^2}} }}{{\sqrt 2 }} + \dfrac{{\sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} }}{{\sqrt 2 }} = \dfrac{{\left| {\sqrt 7 – 1} \right|}}{{\sqrt 2 }} + \dfrac{{\left| {\sqrt 7 + 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 7 – 1}}{{\sqrt 2 }} + \dfrac{{\sqrt 7 + 1}}{{\sqrt 2 }} = \dfrac{{2\sqrt 7 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 7 = \sqrt {14} .\end{array}\)
\(\begin{array}{l}d)\;\;\sqrt {5\sqrt 3 + 5\sqrt {48 – 10\sqrt {7 + 4\sqrt 3 } } } \\ = \sqrt {5\sqrt 3 + 5\sqrt {48 – 10\sqrt {{2^2} + 2.2\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} } } \\ = \sqrt {5\sqrt 3 + 5\sqrt {48 – 10\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} } } = \sqrt {5\sqrt 3 + 5\sqrt {48 – 10.\left| {2 + \sqrt 3 } \right|} } \\ = \sqrt {5\sqrt 3 + 5\sqrt {48 – 10\left( {2 + \sqrt 3 } \right)} } = \sqrt {5\sqrt 3 + 5\sqrt {48 – 20 – 10\sqrt 3 } } \\ = \sqrt {5\sqrt 3 + 5\sqrt {28 – 10\sqrt 3 } } = \sqrt {5\sqrt 3 + 5\sqrt {{5^2} – 2.5.\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} } \\ = \sqrt {5\sqrt 3 + 5\sqrt {{{\left( {5 – \sqrt 3 } \right)}^2}} } = \sqrt {5\sqrt 3 + 5.\left| {5 – \sqrt 3 } \right|} = \sqrt {5\sqrt 3 + 5\left( {5 – \sqrt 3 } \right)} \\ = \sqrt {5\sqrt 3 + 25 – 5\sqrt 3 } = \sqrt {25} = 5.\end{array}\)
\(\begin{array}{l}e)\;\left( {x – 4} \right)\sqrt {16 – 8x + {x^2}} = \left( {x – 4} \right)\sqrt {{{\left( {4 – x} \right)}^2}} \\ = \left( {x – 4} \right).\left| {4 – x} \right| = \left( {x – 4} \right)\left( {x – 4} \right)\;\;\;\left( {do\;\;x \ge 4 \Rightarrow \left| {4 – x} \right| = x – 4} \right)\\ = {\left( {x – 4} \right)^2}.\end{array}\)
\(\begin{array}{l}f)\;\;\left( {2x – 5} \right)\sqrt {\dfrac{2}{{{{\left( {2x – 5} \right)}^2}}}} = \dfrac{{\sqrt 2 .\left( {2x – 5} \right)}}{{\left| {2x – 5} \right|}}\\ = \left\{ \begin{array}{l}\dfrac{{\sqrt 2 \left( {2x – 5} \right)}}{{2x – 5}}\;\;\;khi\;\;2x – 5 > 0\\ – \dfrac{{\sqrt 2 \left( {2x – 5} \right)}}{{2x – 5}}\;\;\;khi\;\;2x – 5 < 0\end{array} \right.\\ = \left\{ \begin{array}{l}\sqrt 2 \;\;\;\;khi\;\;x > \dfrac{5}{2}\\ – \sqrt 2 \;\;\;khi\;\;x < \dfrac{5}{2}\end{array} \right..\end{array}\)
\(\begin{array}{l}g)\;\sqrt {x – 4\sqrt {x – 4} } = \sqrt {x – 4 – 2.2\sqrt {x – 4} + {2^2}} \\ = \sqrt {{{\left( {\sqrt {x – 4} + 2} \right)}^2}} = \left| {\sqrt {x – 4} + 2} \right|\\ = \sqrt {x – 4} + 2\;\;\left( {do\;\;\sqrt {x – 4} + 2 > 0\;\;\sqrt x \ge 4} \right).\end{array}\)