Thử tài bạn 8 trang 28 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Rút gọn :

a) \(2\sqrt {28} + 3\sqrt {63} - 2\sqrt {112} - \sqrt {175} \); 

b) \(\sqrt 2 .\sqrt {7 + 3\sqrt 5 } - \dfrac{4}{{\sqrt 5 - 1}}\);

c) \(\dfrac{1}{{2\sqrt 3 + 3}} - \dfrac{1}{{2\sqrt 3 - 3}}\);

d) \(\sqrt {27b} + 12\sqrt {\dfrac{1}{3}b} - \sqrt {48b} \left( {b > 0} \right)\).

Lời giải chi tiết

\(\begin{array}{l}a)\;2\sqrt {28} + 3\sqrt {63} - 2\sqrt {112} - \sqrt {175} \\ = 2\sqrt {{2^2}.7} + 3\sqrt {{3^2}.7} - 2\sqrt {{4^2}.7} - \sqrt {{5^2}.7} \\ = 2.2\sqrt 7 + 3.3\sqrt 7 - 2.4\sqrt 7 - 5\sqrt 7 \\ = 4\sqrt 7 + 9\sqrt 7 - 8\sqrt 7 - 5\sqrt 7 = 0.\end{array}\)

\(\begin{array}{l}b)\;\sqrt 2 .\sqrt {7 + 3\sqrt 5 } - \dfrac{4}{{\sqrt 5 - 1}}\\ = \sqrt {14 + 6\sqrt 5 } - \dfrac{{4\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}}\\ = \sqrt {{3^2} + 2.3.\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} - \dfrac{{4\left( {\sqrt 5 + 1} \right)}}{{5 - 1}}\\ = \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} - \left( {\sqrt 5 + 1} \right)\\ = \left| {3 + \sqrt 5 } \right| - \sqrt 5 + 1\\ = 3 + \sqrt 5 - \sqrt 5 + 1 = 4.\end{array}\)

\(\begin{array}{l}c)\;\;\dfrac{1}{{2\sqrt 3 + 3}} - \dfrac{1}{{2\sqrt 3 - 3}}\\ = \dfrac{{2\sqrt 3 - 3}}{{\left( {2\sqrt 3 + 3} \right)\left( {2\sqrt 3 - 3} \right)}} - \dfrac{{2\sqrt 3 + 3}}{{\left( {2\sqrt 3 + 3} \right)\left( {2\sqrt 3 - 3} \right)}}\\ = \dfrac{{2\sqrt 3 - 3 - 2\sqrt 3 - 3}}{{{{\left( {2\sqrt 3 } \right)}^2} - 9}}\\ = \dfrac{{ - 6}}{{12 - 9}} = \dfrac{{ - 6}}{3} = - 2.\end{array}\)

\(\begin{array}{l}d)\;\;\sqrt {27b} + 12\sqrt {\dfrac{1}{3}b} - \sqrt {48b} \left( {b > 0} \right)\\ = \sqrt {{3^2}.3b} + 12.\dfrac{{\sqrt {3b} }}{3} - \sqrt {{4^2}.3b} \\ = 3\sqrt {3b} + 4\sqrt {3b} - 4\sqrt {3b} = 3\sqrt {3b} .\end{array}\)

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