Đáp án:
B3:
12) \(x = \dfrac{{11}}{2}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)DK:x \ge \dfrac{1}{2}\\
\sqrt {2x - 1} = \sqrt 5 \\
\to 2x - 1 = 5\\
\to x = 3\\
5)\sqrt 3 {x^2} = \sqrt {12} \\
\to {x^2} = 2\\
\to \left[ \begin{array}{l}
x = \sqrt 2 \\
x = - \sqrt 2
\end{array} \right.\\
9)\sqrt {4{x^2}} = 6\\
\to \left| {2x} \right| = 6\\
\to \left[ \begin{array}{l}
2x = 6\\
2x = - 6
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x = - 3
\end{array} \right.\\
2)DK:x \ge 5\\
\sqrt {x - 5} = 3\\
\to x - 5 = 9\\
\to x = 14\\
6)\left| {x - 3} \right| = 9\\
\to \left[ \begin{array}{l}
x - 3 = 9\\
x - 3 = - 9
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 12\\
x = - 7
\end{array} \right.\\
10)2\left| {1 - x} \right| = 6\\
\to \left| {1 - x} \right| = 3\\
\to \left[ \begin{array}{l}
1 - x = 3\\
1 - x = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 2\\
x = 4
\end{array} \right.\\
3)DK:x \ge 1\\
3\sqrt {x - 1} = 21\\
\to \sqrt {x - 1} = 7\\
\to x - 1 = 49\\
\to x = 50\\
7)\sqrt {{{\left( {2x + 1} \right)}^2}} = 6\\
\to \left| {2x + 1} \right| = 6\\
\to \left[ \begin{array}{l}
2x + 1 = 6\\
2x + 1 = - 6
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = - \dfrac{7}{2}
\end{array} \right.\\
11)x + 1 = {2^3}\\
\to x = 7\\
4)\sqrt 2 x = \sqrt {50} \\
\to x = 5\\
8)\left| {2x - 1} \right| = 3\\
\to \left[ \begin{array}{l}
2x - 1 = 3\\
2x - 1 = - 3
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.\\
12)3 - 2x = {\left( { - 2} \right)^3}\\
\to 3 - 2x = - 8\\
\to 11 = 2x\\
\to x = \dfrac{{11}}{2}\\
B4:\\
a)DK:x \ge 0;x \ne 1\\
A = \left[ {1 + \dfrac{{\sqrt x \left( {\sqrt x + 1} \right)}}{{\sqrt x + 1}}} \right].\left[ {1 - \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x - 1}}} \right]\\
= \left( {1 + \sqrt x } \right)\left( {1 - \sqrt x } \right) = 1 - x\\
b)Do:x \ge 0\\
\to - x \le 0\\
\to 1 - x \le 1\\
\to Max = 1\\
\Leftrightarrow x = 0
\end{array}\)
