Đáp án:
Giải thích các bước giải:
$1)\\a)\sqrt 2 .\sqrt 8 = \sqrt {16} = 4\\ b)\dfrac{{\sqrt 3 }}{{\sqrt {27} }} = \sqrt {\dfrac{1}{9}} = \dfrac{1}{3}\\ c)\sqrt {0,64{a^2}} = 0,8.a\left( {a > 0} \right)\\ d)\sqrt {0,09.{{\left( {a - 1} \right)}^2}} = 0,3.\left( {1 - a} \right)\left( {a < 1} \right) = 0,3 - 0,3a$
$2)\\a)DKXĐ:x \ge 0\\ \sqrt {5x} = 10 \Leftrightarrow 5x = 100 \Leftrightarrow x =20\left( {tmdk} \right)\\ b)\sqrt {6\left( {x - 3} \right)} = 18 \Leftrightarrow 6\left( {x - 3} \right) = {18^2} \Leftrightarrow x - 3 = 54 \Leftrightarrow x = 57\\ c)DK:x \ge 5\\ \sqrt {x - 5} - 2 = 18 \Leftrightarrow \sqrt {x - 5} = 20 \Leftrightarrow x - 5 = 400 \Leftrightarrow x = 405\left( {tm} \right)\\ d)DK:x \ge 5\\ 4 - \sqrt {x - 5} = 8 \Leftrightarrow \sqrt {x - 5} = - 4\left( {ktm} \right)\Leftrightarrow x \in \emptyset$
$3)\\a)DK:x \ge \dfrac{5}{3}\\ \sqrt x = \sqrt {3x - 5} \Leftrightarrow x = 3x - 5 \Leftrightarrow 2x = 5 \Leftrightarrow x = \dfrac{5}{2}\left( {tm} \right)\\ \\ b)Dk:\left\{ \begin{array}{l} 5 - x \ge 0\\ 3x + 2 \ge 0 \end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l} x \le 5\\ x \ge \dfrac{{ - 2}}{3} \end{array} \right.\\ \Leftrightarrow - \dfrac{2}{3} \le x \le 5\\ \sqrt {5 - x} = \sqrt {3x + 2} \Leftrightarrow 5 - x = 3x + 2 \Leftrightarrow 3x + x = 5 - 2\\ \Leftrightarrow 4x = 3 \Leftrightarrow x = \dfrac{3}{4}\left( {tm} \right)\\ c)DKXĐ:x \ge 2\\ \sqrt {4x + 4} = \sqrt {x - 2} \Leftrightarrow 4x + 4 = x - 2 \Leftrightarrow 3x = - 6 \Leftrightarrow x = - 2\left( {ktm} \right)\\\Leftrightarrow x \in \emptyset \\ d)Dk:x \ge 5\\ \sqrt {x + 9} = \sqrt {x - 5} \Leftrightarrow x + 9 = x - 5 \Leftrightarrow 9 = - 5\left( {ktm} \right) \Leftrightarrow,x \in \emptyset$
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