Đáp án:
a. x=-4
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge - 5\\
2\sqrt {x + 5} + \sqrt {x + 5} + \dfrac{1}{3}.3\sqrt {x + 5} = 4\\
\to \left( {2 + 1 + 1} \right)\sqrt {x + 5} = 4\\
\to 4\sqrt {x + 5} = 4\\
\to \sqrt {x + 5} = 1\\
\to x + 5 = 1\\
\to x = - 4\left( {TM} \right)\\
b.DK:x \ge 1\\
6\sqrt {x - 1} - 3\sqrt {x - 1} - 2\sqrt {x - 1} = 16 - \sqrt {x - 1} \\
\to \left( {6 - 3 - 2 + 1} \right)\sqrt {x - 1} = 16\\
\to 2\sqrt {x - 1} = 16\\
\to \sqrt {x - 1} = 8\\
\to x - 1 = 64\\
\to x = 65\left( {TM} \right)\\
c.DK:x \ge - 1\\
\sqrt {x + 7} = 4 - \sqrt {x + 1} \\
\to x + 7 = 16 - 8\sqrt {x + 1} + x + 1\\
\to 8\sqrt {x + 1} = 10\\
\to \sqrt {x + 1} = \dfrac{5}{4}\\
\to x + 1 = \dfrac{{25}}{{16}}\\
\to x = \dfrac{9}{{16}}\left( {TM} \right)\\
d.\sqrt {{{\left( {x + 1} \right)}^2}} + \sqrt {{{\left( {x - 3} \right)}^2}} = 9\\
\to \left| {x + 1} \right| + \left| {x - 3} \right| = 9\\
\to \left[ \begin{array}{l}
x + 1 + x - 3 = 9\left( {DK:x \ge 3} \right)\\
x + 1 - x + 3 = 9\left( {DK: - 1 \le x < 3} \right)\\
- x - 1 - x + 3 = 9\left( {DK:x < - 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
2x = 11\\
4 = 9\left( l \right)\\
- 2x = 7
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{11}}{2}\\
x = - \dfrac{7}{2}
\end{array} \right.\left( {TM} \right)\\
e.DK:x \ge 1\\
\sqrt {x + 2\sqrt {x - 1} } + \sqrt {x - 2\sqrt {x - 1} } = 2\\
\to \sqrt {x - 1 + 2\sqrt {x - 1} .1 + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} .1 + 1} = 2\\
\to \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} = 2\\
\to \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right| = 2\\
\to \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1 = 2\left( {DK:x \ge 2} \right)\\
\to 2\sqrt {x - 1} = 2\\
\to \sqrt {x - 1} = 1\\
\to x - 1 = 1\\
\to x = 2\left( {TM} \right)\\
g.\left( {\sqrt[3]{x} - 5} \right)\left( {\sqrt[3]{x} + 4} \right) = 0\\
\to \left[ \begin{array}{l}
x = {5^3}\\
x = - {4^3}
\end{array} \right. \to \left[ \begin{array}{l}
x = 125\\
x = - 64
\end{array} \right.
\end{array}\)
