Bài 2 trang 19 Tài liệu dạy – học Toán 9 tập 1

Đề bài

 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 \(x \ge 4\).

Lời giải chi tiết

\(\begin{array}{l}a)\;\sqrt {5 + 2\sqrt 6 }  - \sqrt {5 - 2\sqrt 6 }  \\= \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 3 .\sqrt 2  + {{\left( {\sqrt 2 } \right)}^2}} \\ \;- \sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 3 .\sqrt 2  + {{\left( {\sqrt 2 } \right)}^2}} \\ = \sqrt {{{\left( {\sqrt 3  + \sqrt 2 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 3  - \sqrt 2 } \right)}^2}}\\  = \left| {\sqrt 3  + \sqrt 2 } \right| - \left| {\sqrt 3  - \sqrt 2 } \right|\\ = \sqrt 3  + \sqrt 2  - \sqrt 3  + \sqrt 2  \\= 2\sqrt 2 \;\;\;\left( {do\;\;\sqrt 3  > \sqrt 2 } \right).\end{array}\)

\(\begin{array}{l}b)\;\sqrt {8 + \sqrt {60} }  - \sqrt {8 - 2\sqrt {15} } \\ = \sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2.\sqrt 5 .\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}} \\\; - \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 5 .\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}} \\ = \sqrt {{{\left( {\sqrt 5  + \sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 5  - \sqrt 3 } \right)}^2}}  \\= \left| {\sqrt 5  + \sqrt 3 } \right| - \left| {\sqrt 5  - \sqrt 3 } \right|\\ = \sqrt 5  + \sqrt 3  - \sqrt 5  + \sqrt 3\\  = 2\sqrt 3 \;\;\left( {do\;\;\sqrt 5  > \sqrt 3 } \right).\end{array}\)

\(\begin{array}{l}c)\;\sqrt {4 - \sqrt 7 }  + \sqrt {4 + \sqrt 7 }  \\= \dfrac{{\sqrt 2 .\sqrt {4 - \sqrt 7 } }}{{\sqrt 2 }} + \dfrac{{\sqrt 2 .\sqrt {4 + \sqrt 7 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {8 - 2\sqrt 7 } }}{{\sqrt 2 }} + \dfrac{{\sqrt {8 + 2\sqrt 7 } }}{{\sqrt 2 }} \\= \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} - 2\sqrt 7  + 1} }}{{\sqrt 2 }} + \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7  + 1} }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left( {\sqrt 7  - 1} \right)}^2}} }}{{\sqrt 2 }} + \dfrac{{\sqrt {{{\left( {\sqrt 7  + 1} \right)}^2}} }}{{\sqrt 2 }} \\= \dfrac{{\left| {\sqrt 7  - 1} \right|}}{{\sqrt 2 }} + \dfrac{{\left| {\sqrt 7  + 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 7  - 1}}{{\sqrt 2 }} + \dfrac{{\sqrt 7  + 1}}{{\sqrt 2 }} \\= \dfrac{{2\sqrt 7 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 7  = \sqrt {14} .\end{array}\)

\(\begin{array}{l}d)\;\;\sqrt {5\sqrt 3  + 5\sqrt {48 - 10\sqrt {7 + 4\sqrt 3 } } } \\ = \sqrt {5\sqrt 3  + 5\sqrt {48 - 10\sqrt {{2^2} + 2.2\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}} } } \\ = \sqrt {5\sqrt 3  + 5\sqrt {48 - 10\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} } }  \\= \sqrt {5\sqrt 3  + 5\sqrt {48 - 10.\left| {2 + \sqrt 3 } \right|} } \\ = \sqrt {5\sqrt 3  + 5\sqrt {48 - 10\left( {2 + \sqrt 3 } \right)} }  \\= \sqrt {5\sqrt 3  + 5\sqrt {48 - 20 - 10\sqrt 3 } } \\ = \sqrt {5\sqrt 3  + 5\sqrt {28 - 10\sqrt 3 } }  \\= \sqrt {5\sqrt 3  + 5\sqrt {{5^2} - 2.5.\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}} } \\ = \sqrt {5\sqrt 3  + 5\sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} } \\ = \sqrt {5\sqrt 3  + 5.\left| {5 - \sqrt 3 } \right|}  \\= \sqrt {5\sqrt 3  + 5\left( {5 - \sqrt 3 } \right)} \\ = \sqrt {5\sqrt 3  + 25 - 5\sqrt 3 }  = \sqrt {25}  = 5.\end{array}\)

\(\begin{array}{l}e)\;\left( {x - 4} \right)\sqrt {16 - 8x + {x^2}}  \\= \left( {x - 4} \right)\sqrt {{{\left( {4 - x} \right)}^2}} \\ = \left( {x - 4} \right).\left| {4 - x} \right| \\= \left( {x - 4} \right)\left( {x - 4} \right)\;\;\;\left( {do\;\;x \ge 4 \Rightarrow \left| {4 - x} \right| = x - 4} \right)\\ = {\left( {x - 4} \right)^2}.\end{array}\)

\(\begin{array}{l}f)\;\;\left( {2x - 5} \right)\sqrt {\dfrac{2}{{{{\left( {2x - 5} \right)}^2}}}} \\ = \dfrac{{\sqrt 2 .\left( {2x - 5} \right)}}{{\left| {2x - 5} \right|}}\\ = \left\{ \begin{array}{l}\dfrac{{\sqrt 2 \left( {2x - 5} \right)}}{{2x - 5}}\;\;\;khi\;\;2x - 5 > 0\\ - \dfrac{{\sqrt 2 \left( {2x - 5} \right)}}{{2x - 5}}\;\;\;khi\;\;2x - 5 < 0\end{array} \right.\\ = \left\{ \begin{array}{l}\sqrt 2 \;\;\;\;khi\;\;x > \dfrac{5}{2}\\ - \sqrt 2 \;\;\;khi\;\;x < \dfrac{5}{2}\end{array} \right..\end{array}\)

\(\begin{array}{l}g)\;\sqrt {x - 4\sqrt {x - 4} }  = \sqrt {x - 4 - 2.2\sqrt {x - 4}  + {2^2}} \\ = \sqrt {{{\left( {\sqrt {x - 4}  + 2} \right)}^2}}  \\= \left| {\sqrt {x - 4}  + 2} \right|\\ = \sqrt {x - 4}  + 2\;\;\left( {do\;\;\sqrt {x - 4}  + 2 > 0\;\;\sqrt x  \ge 4} \right).\end{array}\)