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