Đáp án:
Câu 5:
b.\(x = 6 + 2\sqrt 7 \)
Giải thích các bước giải:
\(\begin{array}{l}
C4:\\
a.DK:a > 0;b > 0;a \ne b\\
b.A = \dfrac{1}{{\sqrt b \left( {\sqrt b - \sqrt a } \right)}} + \dfrac{1}{{\sqrt a \left( {\sqrt a + \sqrt b } \right)}}\\
= \dfrac{{a + \sqrt {ab} + b - \sqrt {ab} }}{{\sqrt {ab} \left( {b - a} \right)}} = \dfrac{{a + b}}{{\sqrt {ab} \left( {b - a} \right)}}\\
C5:\\
a.DK:x \ge \dfrac{1}{2}\\
\sqrt {2x - 1} = - 3\left( {ld} \right)\\
Do:\sqrt {2x - 1} \ge 0\forall x \ge \dfrac{1}{2}\\
b.DK:x \ge \dfrac{1}{2}\\
x - 2\sqrt {2x - 1} = 2\\
\to x - 2 = 2\sqrt {2x - 1} \left( {DK:x \ge 2} \right)\\
\to {x^2} - 4x + 4 = 4\left( {2x - 1} \right)\\
\to {x^2} - 12x + 8 = 0\\
\to \left[ \begin{array}{l}
x = 6 + 2\sqrt 7 \\
x = 6 + 2\sqrt 7 \left( l \right)
\end{array} \right.
\end{array}\)
