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 \[x \ge 4\].
Phương pháp giải - Xem chi tiết
+] Sử dụng công thức; \[\sqrt {{A^2}} = \left| A \right| = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ - A\;\;khi\;A < 0\end{array} \right..\]
Lời giải chi tiết
\[\begin{array}{l}a]\;\sqrt {5 + 2\sqrt 6 } - \sqrt {5 - 2\sqrt 6 } \\= \sqrt {{{\left[ {\sqrt 3 } \right]}^2} + 2.\sqrt 3 .\sqrt 2 + {{\left[ {\sqrt 2 } \right]}^2}} \\ \;- \sqrt {{{\left[ {\sqrt 3 } \right]}^2} - 2.\sqrt 3 .\sqrt 2 + {{\left[ {\sqrt 2 } \right]}^2}} \\ = \sqrt {{{\left[ {\sqrt 3 + \sqrt 2 } \right]}^2}} - \sqrt {{{\left[ {\sqrt 3 - \sqrt 2 } \right]}^2}}\\ = \left| {\sqrt 3 + \sqrt 2 } \right| - \left| {\sqrt 3 - \sqrt 2 } \right|\\ = \sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 \\= 2\sqrt 2 .\end{array}\]
\[\begin{array}{l}b]\;\sqrt {8 + \sqrt {60} } - \sqrt {8 - 2\sqrt {15} } \\ = \sqrt {{{\left[ {\sqrt 5 } \right]}^2} + 2.\sqrt 5 .\sqrt 3 + {{\left[ {\sqrt 3 } \right]}^2}} \\\; - \sqrt {{{\left[ {\sqrt 5 } \right]}^2} - 2.\sqrt 5 .\sqrt 3 + {{\left[ {\sqrt 3 } \right]}^2}} \\ = \sqrt {{{\left[ {\sqrt 5 + \sqrt 3 } \right]}^2}} - \sqrt {{{\left[ {\sqrt 5 - \sqrt 3 } \right]}^2}} \\= \left| {\sqrt 5 + \sqrt 3 } \right| - \left| {\sqrt 5 - \sqrt 3 } \right|\\ = \sqrt 5 + \sqrt 3 - \sqrt 5 + \sqrt 3\\ = 2\sqrt 3 .\end{array}\]
\[\begin{array}{l}c]\;\sqrt {4 - \sqrt 7 } + \sqrt {4 + \sqrt 7 } \\= \dfrac{{\sqrt 2 .\sqrt {4 - \sqrt 7 } }}{{\sqrt 2 }} + \dfrac{{\sqrt 2 .\sqrt {4 + \sqrt 7 } }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {8 - 2\sqrt 7 } }}{{\sqrt 2 }} + \dfrac{{\sqrt {8 + 2\sqrt 7 } }}{{\sqrt 2 }} \\= \dfrac{{\sqrt {{{\left[ {\sqrt 7 } \right]}^2} - 2\sqrt 7 + 1} }}{{\sqrt 2 }} + \dfrac{{\sqrt {{{\left[ {\sqrt 7 } \right]}^2} + 2\sqrt 7 + 1} }}{{\sqrt 2 }}\\ = \dfrac{{\sqrt {{{\left[ {\sqrt 7 - 1} \right]}^2}} }}{{\sqrt 2 }} + \dfrac{{\sqrt {{{\left[ {\sqrt 7 + 1} \right]}^2}} }}{{\sqrt 2 }} \\= \dfrac{{\left| {\sqrt 7 - 1} \right|}}{{\sqrt 2 }} + \dfrac{{\left| {\sqrt 7 + 1} \right|}}{{\sqrt 2 }}\\ = \dfrac{{\sqrt 7 - 1}}{{\sqrt 2 }} + \dfrac{{\sqrt 7 + 1}}{{\sqrt 2 }} \\= \dfrac{{2\sqrt 7 }}{{\sqrt 2 }} = \sqrt 2 .\sqrt 7 = \sqrt {14} .\end{array}\]
\[\begin{array}{l}d]\;\;\sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {7 + 4\sqrt 3 } } } \\ = \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {{2^2} + 2.2\sqrt 3 + {{\left[ {\sqrt 3 } \right]}^2}} } } \\ = \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {{{\left[ {2 + \sqrt 3 } \right]}^2}} } } \\= \sqrt {5\sqrt 3 + 5\sqrt {48 - 10.\left| {2 + \sqrt 3 } \right|} } \\ = \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\left[ {2 + \sqrt 3 } \right]} } \\= \sqrt {5\sqrt 3 + 5\sqrt {48 - 20 - 10\sqrt 3 } } \\ = \sqrt {5\sqrt 3 + 5\sqrt {28 - 10\sqrt 3 } } \\= \sqrt {5\sqrt 3 + 5\sqrt {{5^2} - 2.5.\sqrt 3 + {{\left[ {\sqrt 3 } \right]}^2}} } \\ = \sqrt {5\sqrt 3 + 5\sqrt {{{\left[ {5 - \sqrt 3 } \right]}^2}} } \\ = \sqrt {5\sqrt 3 + 5.\left| {5 - \sqrt 3 } \right|} \\= \sqrt {5\sqrt 3 + 5\left[ {5 - \sqrt 3 } \right]} \\ = \sqrt {5\sqrt 3 + 25 - 5\sqrt 3 } = \sqrt {25} = 5.\end{array}\]
\[\begin{array}{l}e]\;\left[ {x - 4} \right]\sqrt {16 - 8x + {x^2}} \\= \left[ {x - 4} \right]\sqrt {{{\left[ {4 - x} \right]}^2}} \\ = \left[ {x - 4} \right].\left| {4 - x} \right| \\= \left[ {x - 4} \right]\left[ {x - 4} \right]\;\;\;\left[ {do\;\;x \ge 4 \Rightarrow \left| {4 - x} \right| = x - 4} \right]\\ = {\left[ {x - 4} \right]^2}.\end{array}\]
\[\begin{array}{l}f]\;\;\left[ {2x - 5} \right]\sqrt {\dfrac{2}{{{{\left[ {2x - 5} \right]}^2}}}} \\ = \dfrac{{\sqrt 2 .\left[ {2x - 5} \right]}}{{\left| {2x - 5} \right|}}\\ = \left\{ \begin{array}{l}\dfrac{{\sqrt 2 \left[ {2x - 5} \right]}}{{2x - 5}}\;\;\;khi\;\;2x - 5 > 0\\ - \dfrac{{\sqrt 2 \left[ {2x - 5} \right]}}{{2x - 5}}\;\;\;khi\;\;2x - 5 < 0\end{array} \right.\\ = \left\{ \begin{array}{l}\sqrt 2 \;\;\;\;khi\;\;x > \dfrac{5}{2}\\ - \sqrt 2 \;\;\;khi\;\;x < \dfrac{5}{2}\end{array} \right..\end{array}\]
\[\begin{array}{l}g]\;\sqrt {x - 4\sqrt {x - 4} } = \sqrt {x - 4 - 2.2\sqrt {x - 4} + {2^2}} \\ = \sqrt {{{\left[ {\sqrt {x - 4} + 2} \right]}^2}} \\= \left| {\sqrt {x - 4} + 2} \right|\\ = \sqrt {x - 4} + 2\;\;\left[ {do\;\;\sqrt {x - 4} + 2 > 0\;\;\sqrt x \ge 4} \right].\end{array}\]