\(\begin{array}{l}c)\;2\sqrt 5 - 2\sqrt {108} + 3\sqrt {20} + 5\sqrt {27} \\ = 2\sqrt 5 - 2\sqrt {{6^2}.3} + 3\sqrt {{2^2}.5} + 5\sqrt {{3^2}.3} \\ = 2\sqrt 5 - 2.6\sqrt 3 + 3.2\sqrt 5 + 5.3\sqrt 3 \\ = 2\sqrt 5 - 12\sqrt 3 + 6\sqrt 5 + 15\sqrt 3 \\ = 8\sqrt 5 + 3\sqrt 3 .\end{array}\) \(\begin{array}{l}d)\;\;8\sqrt {{x^3}{y^2}} - 3y\sqrt {{x^3}} \left( {x \ge 0,y \ge 0} \right)\\ = 8\sqrt {{{\left( {xy} \right)}^2}x} - 3y\sqrt {{x^2}.x} \\ = 8\left| {xy} \right|.\sqrt x - 3y\left| x \right|\sqrt x \\ = 8xy\sqrt x - 3xy\sqrt x \;\;\left( {do\;\;x \ge 0,y \ge 0} \right)\\ = 5xy\sqrt x .\end{array}\) Đề bài Rút gọn biểu thức : a) \(\sqrt {18} - 2\sqrt {32} + 3\sqrt {50} - 4\sqrt 8 \); b) \(3\sqrt {27} + \sqrt {75} - 3\sqrt {48} - 4\sqrt {12} \); c) \(2\sqrt 5 - 2\sqrt {108} + 3\sqrt {20} + 5\sqrt {27} \); d) \(8\sqrt {{x^3}{y^2}} - 3y\sqrt {{x^3}} \left( {x \ge 0,y \ge 0} \right)\). Phương pháp giải - Xem chi tiết +) Sử dụng công thức: \(\sqrt {{A^2}B} = \left| A \right|\sqrt B = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ - A\;\;\;khi\;\;A < 0\end{array} \right..\) Lời giải chi tiết \(\begin{array}{l}a)\;\;\sqrt {18} - 2\sqrt {32} + 3\sqrt {50} - 4\sqrt 8 \\ = \sqrt {{3^2}.2} - 2\sqrt {{4^2}.2} + 3\sqrt {{5^2}.2} - 4\sqrt {{2^2}.2} \\ = 3\sqrt 2 - 2.4\sqrt 2 + 3.5\sqrt 2 - 4.2\sqrt 2 \\ = 3\sqrt 2 - 8\sqrt 2 + 15\sqrt 2 - 8\sqrt 2 \\ = 2\sqrt 2 .\end{array}\) \(\begin{array}{l}b)\;3\sqrt {27} + \sqrt {75} - 3\sqrt {48} - 4\sqrt {12} \\ = 3\sqrt {{3^2}.3} + \sqrt {{5^2}.3} - 3\sqrt {{4^2}.3} - 4\sqrt {{2^2}.3} \\ = 3.3\sqrt 3 + 5\sqrt 3 - 3.4\sqrt 3 - 4.2\sqrt 3 \\ = 9\sqrt 3 + 5\sqrt 3 - 12\sqrt 3 - 8\sqrt 3 \\ = - 6\sqrt 3 .\end{array}\) \(\begin{array}{l}c)\;2\sqrt 5 - 2\sqrt {108} + 3\sqrt {20} + 5\sqrt {27} \\ = 2\sqrt 5 - 2\sqrt {{6^2}.3} + 3\sqrt {{2^2}.5} + 5\sqrt {{3^2}.3} \\ = 2\sqrt 5 - 2.6\sqrt 3 + 3.2\sqrt 5 + 5.3\sqrt 3 \\ = 2\sqrt 5 - 12\sqrt 3 + 6\sqrt 5 + 15\sqrt 3 \\ = 8\sqrt 5 + 3\sqrt 3 .\end{array}\) \(\begin{array}{l}d)\;\;8\sqrt {{x^3}{y^2}} - 3y\sqrt {{x^3}} \left( {x \ge 0,y \ge 0} \right)\\ = 8\sqrt {{{\left( {xy} \right)}^2}x} - 3y\sqrt {{x^2}.x} \\ = 8\left| {xy} \right|.\sqrt x - 3y\left| x \right|\sqrt x \\ = 8xy\sqrt x - 3xy\sqrt x \;\;\left( {do\;\;x \ge 0,y \ge 0} \right)\\ = 5xy\sqrt x .\end{array}\)
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