Đáp án:
10) \(x \in \emptyset \)
Giải thích các bước giải:
\(\begin{array}{l}
6)DK:0 \le x \le 1\\
\dfrac{5}{3}.3.\sqrt {5x} - 5\sqrt {5x} - \dfrac{1}{3}.9\sqrt {5x} - 2\sqrt {16\left( {1 - x} \right)} = 0\\
\to - 3\sqrt {5x} = 2.4\sqrt {1 - x} \\
\to - \dfrac{3}{8}\sqrt {5x} = \sqrt {1 - x} \left( {vô lý} \right)\\
\to x \in \emptyset \\
7)DK:x \ge 5\\
\dfrac{1}{5}.5\sqrt {x - 5} - \dfrac{3}{2}\sqrt {x - 5} + 6\sqrt {x - 5} + 3\sqrt {x - 3} = 0\\
\to \dfrac{{11}}{2}\sqrt {x - 5} + 3\sqrt {x - 3} = 0\\
\to \dfrac{{11}}{2}\sqrt {x - 5} = - 3\sqrt {x - 3} \left( {vô lý} \right)\\
\to x \in \emptyset \\
8)DK:\left[ \begin{array}{l}
x \ge \sqrt 5 \\
x \le 0
\end{array} \right.\\
\sqrt {{x^2} - 2x.\sqrt 5 + 5} + \sqrt {x\left( {x - \sqrt 5 } \right)} = 0\\
\to \sqrt {{{\left( {x - \sqrt 5 } \right)}^2}} + \sqrt {x\left( {x - \sqrt 5 } \right)} = 0\\
\to \sqrt {x - \sqrt 5 } .\left( {\sqrt {x - \sqrt 5 } + \sqrt x } \right) = 0\\
\to \left[ \begin{array}{l}
x - \sqrt 5 = 0\\
\sqrt {x - \sqrt 5 } + \sqrt x = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \sqrt 5 \\
\sqrt {x - \sqrt 5 } = - \sqrt x \left( {vô lý} \right)
\end{array} \right.\\
\to x = \sqrt 5 \\
9)DK:\left[ \begin{array}{l}
x \ge 4\\
x \le - 4
\end{array} \right.\\
\sqrt {x\left( {x + 4} \right)} + \sqrt {\dfrac{{\left( {x - 4} \right)\left( {x + 4} \right)}}{2}} = 0\\
\to \sqrt {x + 4} \left( {\sqrt x + \sqrt {\dfrac{{x - 4}}{2}} } \right) = 0\\
\to x + 4 = 0\left( {do:\sqrt x + \sqrt {\dfrac{{x - 4}}{2}} > 0\forall x \ge 4} \right)\\
\to x = - 4\\
10)DK:x \ge 2\\
\sqrt {x - 1} + \sqrt {\left( {x - 1} \right)\left( {x - 2} \right)} = 0\\
\to \sqrt {x - 1} \left( {1 + \sqrt {x - 2} } \right) = 0\\
\to x - 1 = 0\left( {do:1 + \sqrt {x - 2} > 0\forall x \ge 2} \right)\\
\to x = 1\left( l \right)\\
\to x \in \emptyset
\end{array}\)
