Đáp án:
6) x=24
Giải thích các bước giải:
\(\begin{array}{l}
3)DK:x \ge - 1;x \ne 0\\
\dfrac{{\sqrt {x + 1} - 1 - \sqrt {x + 1} - 1 + 2\left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt {x + 1} + 1} \right)}}{{x + 1 - 1}} = 0\\
\to - 2 + 2\left( {x + 1 - 1} \right) = 0\\
\to 2x = 2\\
\to x = 1\left( {TM} \right)\\
4)\dfrac{{2x\left( {\sqrt 5 + \sqrt 3 } \right)}}{{5 - 3}} - \dfrac{{2x\left( {\sqrt 3 - 1} \right)}}{{3 - 1}} = \sqrt 5 + 1\\
\to x\left( {\sqrt 5 + \sqrt 3 } \right) - x\left( {\sqrt 3 - 1} \right) = \sqrt 5 + 1\\
\to x\sqrt 5 + x\sqrt 3 - x\sqrt 3 + x = \sqrt 5 + 1\\
\to x\left( {\sqrt 5 + 1} \right) = \sqrt 5 + 1\\
\to x = 1\\
5)DK:x \ge 0\\
2x - 7\sqrt x + 5 = 0\\
\to \left( {2\sqrt x - 5} \right)\left( {\sqrt x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{5}{2}\\
\sqrt x = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{25}}{4}\\
x = 1
\end{array} \right.\\
6)DK:x \ge 3\\
x - 3 - 6\sqrt {x - 3} - 7 = 0\\
\to \left( {\sqrt {x - 3} - 7} \right)\left( {\sqrt {x - 3} + 1} \right) = 0\\
\to \sqrt {x - 3} - 7 = 0\left( {do:\sqrt {x - 3} + 1 > 0\forall x \ge 3} \right)\\
\to \sqrt {x - 3} = 7\\
\to x - 3 = 21\\
\to x = 24
\end{array}\)
