Bài 9 trang 30 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Rút gọn :

a) \(4\sqrt {12}  - 3\sqrt {48}  + 5\sqrt {75}  - 2\sqrt {108} \); 

b) \(\sqrt {20}  - 3\sqrt {80}  + 5\sqrt {45}  - 2\sqrt {125} \);

c) \(\sqrt {60}  - 4\sqrt {\dfrac{3}{5}}  + \sqrt {\dfrac{5}{3}} \);

d) \(4\sqrt {2x}  - 7\sqrt {18x}  - 5\sqrt {8x}  + \sqrt {32x} \)\(\;\left( {x \ge 0} \right)\);

e) \(\sqrt {50{x^2}{y^5}}  - 13y\sqrt {2{x^2}{y^3}}  + xy\sqrt {98{y^3}} \)\(\;\left( {x \ge 0,y \ge 0} \right)\).

Lời giải chi tiết

\(\begin{array}{l}a)\;4\sqrt {12}  - 3\sqrt {48}  + 5\sqrt {75}  - 2\sqrt {108} \\ = 4\sqrt {{2^2}.3}  - 3\sqrt {{4^2}.3}  + 5\sqrt {{5^2}.3}  - 2\sqrt {{6^2}.3} \\ = 4.2\sqrt 3  - 3.4\sqrt 3  + 5.5\sqrt 3  - 2.6\sqrt 3 \\ = 9\sqrt 3 .\end{array}\)                         \(\begin{array}{l}b)\;\sqrt {20}  - 3\sqrt {80}  + 5\sqrt {45}  - 2\sqrt {125} \\ = \sqrt {{2^2}.5}  - 3\sqrt {{4^2}.5}  + 5\sqrt {{3^2}.5}  - 2\sqrt {{5^2}.5} \\ = 2\sqrt 5  - 3.4\sqrt 5  + 5.3\sqrt 5  - 2.5\sqrt 5 \\ =  - 5\sqrt 5 .\end{array}\)

\(\begin{array}{l}c)\;\sqrt {60}  - 4\sqrt {\dfrac{3}{5}}  + \sqrt {\dfrac{5}{3}} \\ = \sqrt {{2^2}.15}  - 4\sqrt {\dfrac{{3.5}}{{{5^2}}}}  + \sqrt {\dfrac{{5.3}}{{{3^2}}}} \\ = 2\sqrt {15}  - \dfrac{{4\sqrt {15} }}{5} + \dfrac{{\sqrt {15} }}{3}\\ = \dfrac{{23\sqrt {15} }}{{15}}.\end{array}\)                                                      \(\begin{array}{l}d)\;4\sqrt {2x}  - 7\sqrt {18x}  - 5\sqrt {8x}  + \sqrt {32x} \;\;\left( {x \ge 0} \right)\\ = 4\sqrt {2x}  - 7\sqrt {{3^2}.2x}  - 5\sqrt {{2^2}.2x}  + \sqrt {{4^2}.2x} \\ = 4\sqrt {2x}  - 7.3\sqrt {2x}  - 5.2\sqrt {2x}  + 4\sqrt {2x} \\ =  - 23\sqrt {2x} .\end{array}\)

\(\begin{array}{l}e)\;\sqrt {50{x^2}{y^5}}  - 13y\sqrt {2{x^2}{y^3}}  + xy\sqrt {98{y^3}} \;\;\left( {x \ge 0,\;\;y \ge 0} \right)\\ = \sqrt {{{2.5}^2}{{\left( {x{y^2}} \right)}^2}y}  - 13y\sqrt {2{{\left( {xy} \right)}^2}.y}  + xy\sqrt {{7^2}.2.{y^2}.y} \\ = 5\left| {x{y^2}} \right|\sqrt {2y}  - 13y\left| {xy} \right|\sqrt {2y}  + xy.7\left| y \right|\sqrt {2y} \\ = 5x{y^2}\sqrt {2y}  - 13x{y^2}\sqrt {2y}  + 7x{y^2}\sqrt {2y} \\ =  - x{y^2}\sqrt {2y} .\end{array}\)

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