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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) \(\sqrt {4 + 2\sqrt 3 } – \sqrt {4 – 2\sqrt 3 } = 2\)
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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.
\(\begin{array}{l}a)\;\;\sqrt {9 – 4\sqrt 5 } – \sqrt 5 = – 2\\VT = \sqrt {9 – 4\sqrt 5 } – \sqrt 5 = \sqrt {{2^2} – 2.2.\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} – \sqrt 5 \\ = \sqrt {{{\left( {2 – \sqrt 5 } \right)}^2}} – \sqrt 5 = \left| {2 – \sqrt 5 } \right| – \sqrt 5 \\ = \sqrt 5 – 2 – \sqrt 5 = – 2 = VP\;\;\left( {dpcm} \right).\;\;\end{array}\)\(\begin{array}{l}b)\;\;{\left( {4 – \sqrt 7 } \right)^2} = 23 – 8\sqrt 7 \\VT = {\left( {4 – \sqrt 7 } \right)^2} = {4^2} – 2.4.\sqrt 7 + {\left( {\sqrt 7 } \right)^2}\\ = 16 – 8\sqrt 7 + 7 = 23 – 8\sqrt 7 = VP\;\;\;\left( {dpcm} \right).\end{array}\)
\(\begin{array}{l}c)\;\sqrt {11 – 2\sqrt {10} } – \sqrt {10} = – 1\\VT = \sqrt {11 – 2\sqrt {10} } – \sqrt {10} = \sqrt {{{\left( {\sqrt {10} } \right)}^2} – 2\sqrt {10} + 1} – \sqrt {10} \\ = \sqrt {{{\left( {\sqrt {10} – 1} \right)}^2}} – \sqrt {10} = \sqrt {10} – 1 – \sqrt {10} = – 1 = VP\;\;\left( {dpcm} \right).\end{array}\)
\(\begin{array}{l}d)\;\sqrt {4 + 2\sqrt 3 } – \sqrt {4 – 2\sqrt 3 } = 2\\VT = \sqrt {4 + 2\sqrt 3 } – \sqrt {4 – 2\sqrt 3 } = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2\sqrt 3 + 1} – \sqrt {{{\left( {\sqrt 3 } \right)}^2} – 2\sqrt 3 + 1} \\ = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} = \left| {\sqrt 3 + 1} \right| – \left| {\sqrt 3 – 1} \right|\\ = \sqrt 3 + 1 – \sqrt 3 + 1 = 2 = VP\;\;\;\left( {dpcm} \right).\end{array}\)