Bài 9 trang 18 Tài liệu dạy – học Toán 9 tập 1

Đề bài

 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) \(\sqrt {4 + 2\sqrt 3 }  – \sqrt {4 – 2\sqrt 3 }  = 2\)

Lời giải chi tiết

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

\(\begin{array}{l}c)\;\sqrt {11 – 2\sqrt {10} }  – \sqrt {10}  =  – 1\\VT = \sqrt {11 – 2\sqrt {10} }  – \sqrt {10}  \\= \sqrt {{{\left( {\sqrt {10} } \right)}^2} – 2\sqrt {10}  + 1}  – \sqrt {10} \\ = \sqrt {{{\left( {\sqrt {10}  – 1} \right)}^2}}  – \sqrt {10}\\  = \sqrt {10}  – 1 – \sqrt {10} \\ =  – 1 = VP\;\;\left( {đpcm} \right).\end{array}\)

\(\begin{array}{l}d)\;\sqrt {4 + 2\sqrt 3 }  – \sqrt {4 – 2\sqrt 3 }  = 2\\VT = \sqrt {4 + 2\sqrt 3 }  – \sqrt {4 – 2\sqrt 3 }  \\= \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2\sqrt 3  + 1}  – \sqrt {{{\left( {\sqrt 3 } \right)}^2} – 2\sqrt 3  + 1} \\ = \sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  – \sqrt {{{\left( {\sqrt 3  – 1} \right)}^2}}  \\= \left| {\sqrt 3  + 1} \right| – \left| {\sqrt 3  – 1} \right|\\ = \sqrt 3  + 1 – \sqrt 3  + 1 = 2 = VP\;\;\;\left( {đpcm} \right).\end{array}\)

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