Chứng minh :
a) \(\sqrt {9 - 4\sqrt 5 } - \sqrt 5 = - 2\);
b) \({\left( {4 - \sqrt 7 } \right)^2} = 23 - 8\sqrt 7 \);
c)\(\sqrt {11 - 2\sqrt {10} } - \sqrt {10} = - 1\) ;
d) \(\sqrt {4 + 2\sqrt 3 } - \sqrt {4 - 2\sqrt 3 } = 2\)
Advertisements (Quảng cáo)
Sử dụng công thức: \(\sqrt {{A^2}} = \left| A \right| = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ - A\;\;khi\;\;A < 0\end{array} \right.\)biến đổi vế trái thành biểu thức như vế phải.
\(\begin{array}{l}a)\;\;\sqrt {9 - 4\sqrt 5 } - \sqrt 5 = - 2\\VT = \sqrt {9 - 4\sqrt 5 } - \sqrt 5 = \sqrt {{2^2} - 2.2.\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} - \sqrt 5 \\ = \sqrt {{{\left( {2 - \sqrt 5 } \right)}^2}} - \sqrt 5 = \left| {2 - \sqrt 5 } \right| - \sqrt 5 \\ = \sqrt 5 - 2 - \sqrt 5 = - 2 = VP\;\;\left( {dpcm} \right).\;\;\end{array}\)\(\begin{array}{l}b)\;\;{\left( {4 - \sqrt 7 } \right)^2} = 23 - 8\sqrt 7 \\VT = {\left( {4 - \sqrt 7 } \right)^2} = {4^2} - 2.4.\sqrt 7 + {\left( {\sqrt 7 } \right)^2}\\ = 16 - 8\sqrt 7 + 7 = 23 - 8\sqrt 7 = VP\;\;\;\left( {dpcm} \right).\end{array}\)
\(\begin{array}{l}c)\;\sqrt {11 - 2\sqrt {10} } - \sqrt {10} = - 1\\VT = \sqrt {11 - 2\sqrt {10} } - \sqrt {10} = \sqrt {{{\left( {\sqrt {10} } \right)}^2} - 2\sqrt {10} + 1} - \sqrt {10} \\ = \sqrt {{{\left( {\sqrt {10} - 1} \right)}^2}} - \sqrt {10} = \sqrt {10} - 1 - \sqrt {10} = - 1 = VP\;\;\left( {dpcm} \right).\end{array}\)
\(\begin{array}{l}d)\;\sqrt {4 + 2\sqrt 3 } - \sqrt {4 - 2\sqrt 3 } = 2\\VT = \sqrt {4 + 2\sqrt 3 } - \sqrt {4 - 2\sqrt 3 } = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2\sqrt 3 + 1} - \sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2\sqrt 3 + 1} \\ = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} = \left| {\sqrt 3 + 1} \right| - \left| {\sqrt 3 - 1} \right|\\ = \sqrt 3 + 1 - \sqrt 3 + 1 = 2 = VP\;\;\;\left( {dpcm} \right).\end{array}\)