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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