Đáp án:
c. \(\left[ \begin{array}{l}
x = - 1\\
x = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge 2\\
2x - 4 = \sqrt {x - 2} \\
\to 4{x^2} - 16x + 16 = x - 2\\
\to 4{x^2} - 17x + 18 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{9}{4}\\
x = 2
\end{array} \right.\left( {TM} \right)\\
b.DK:x \ge - 3\\
3\sqrt {x + 3} + 5\sqrt {x + 3} - \dfrac{3}{4}.4.\sqrt {x + 3} = 3\\
\to \left( {3 + 5 - 3} \right)\sqrt {x + 3} = 3\\
\to 5\sqrt {x + 3} = 3\\
\to \sqrt {x + 3} = \dfrac{3}{5}\\
\to x + 3 = \dfrac{9}{{25}}\\
\to x = - \dfrac{{66}}{{25}}\left( {TM} \right)\\
c.\sqrt {{x^2} + x + 1} = 1\\
\to {x^2} + x + 1 = 1\\
\to {x^2} + x = 0\\
\to x\left( {x + 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = - 1\\
x = 0
\end{array} \right.
\end{array}\)
