Giải thích các bước giải:
Bài 1:
Các biểu thức đã cho xác định khi:
\(\begin{array}{l}
a,\\
\left\{ \begin{array}{l}
3x - 2 \ge 0\\
\frac{{ - 4}}{{x - 3}} \ge 0\\
x - 3 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3x - 2 \ge 0\\
x - 3 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge \frac{2}{3}\\
x < 3
\end{array} \right. \Leftrightarrow \frac{2}{3} \le x < 3\\
b,\\
{x^4} + 1 \ge 0,\,\,\,\,\forall x\\
2,\\
a,\\
\sqrt {16{x^2} - 8x + 1} = 9\\
\Leftrightarrow \sqrt {{{\left( {4x - 1} \right)}^2}} = 9\\
\Leftrightarrow \left| {4x - 1} \right| = 9\\
\Leftrightarrow \left[ \begin{array}{l}
4x - 1 = 9\\
4x - 1 = - 9
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{5}{2}\\
x = - 2
\end{array} \right.\\
b,\\
\sqrt x + \sqrt {16x} - \sqrt {4x} = 3\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \sqrt x + \sqrt {{4^2}.x} - \sqrt {{2^2}.x} = 3\\
\Leftrightarrow \sqrt x + 4\sqrt x - 2\sqrt x = 3\\
\Leftrightarrow 3\sqrt x = 3\\
\Leftrightarrow \sqrt x = 1\\
\Leftrightarrow x = 1\\
c,\\
\sqrt {x - 2} + \sqrt {4x - 8} - \sqrt {25x - 50} + 6 = 0\,\,\,\,\,\,\,\,\,\,\left( {x \ge 2} \right)\\
\Leftrightarrow \sqrt {x - 2} + \sqrt {4\left( {x - 2} \right)} - \sqrt {25\left( {x - 2} \right)} + 6 = 0\\
\Leftrightarrow \sqrt {x - 2} + \sqrt {{2^2}.\left( {x - 2} \right)} - \sqrt {{5^2}\left( {x - 2} \right)} + 6 = 0\\
\Leftrightarrow \sqrt {x - 2} + 2\sqrt {x - 2} - 5\sqrt {x - 2} + 6 = 0\\
\Leftrightarrow - 2\sqrt {x - 2} + 6 = 0\\
\Leftrightarrow \sqrt {x - 2} = 3\\
\Leftrightarrow x = 11
\end{array}\)
