Đáp án:
d. \(\left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt {17} }}{2}\\
x = \dfrac{{3 - \sqrt {17} }}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.DK:x \ge - \dfrac{1}{3}\\
a.\sqrt {3x + 1} = 3x - 1\\
\to 3x + 1 = 9{x^2} - 6x + 1\left( {DK:x \ge \dfrac{1}{3}} \right)\\
\to 9{x^2} - 9x = 0\\
\to 9x\left( {x - 1} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\left( l \right)\\
x = 1\left( {TM} \right)
\end{array} \right.\\
b.DK:x \ge - 1\\
Pt \to 2 + \sqrt {3x + 5} = x + 1\\
\to \sqrt {3x + 5} = x - 1\\
\to 3x + 5 = {x^2} - 2x + 1\left( {DK:x \ge 1} \right)\\
\to {x^2} - 5x - 4 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{5 + \sqrt {41} }}{2}\left( {TM} \right)\\
x = \dfrac{{5 - \sqrt {41} }}{2}\left( l \right)
\end{array} \right.\\
c.DK:x \ge \dfrac{1}{2}\\
Pt \to 5{x^2} = 4{x^2} - 4x + 1\\
\to {x^2} + 4x - 1 = 0\\
\to \left[ \begin{array}{l}
x = - 2 + \sqrt 5 \\
x = - 2 - \sqrt 5
\end{array} \right.\left( l \right)\\
\to x \in \emptyset \\
d.\sqrt {{{\left( {x - 1} \right)}^2}} + \sqrt {{{\left( {x - 2} \right)}^2}} = 3\\
\to \left| {x - 1} \right| + \left| {x - 2} \right| = 3\\
\to {x^2} - 2x + 1 + {x^2} - 4x + 4 = 9\\
\to 2{x^2} - 6x - 4 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{3 + \sqrt {17} }}{2}\\
x = \dfrac{{3 - \sqrt {17} }}{2}
\end{array} \right.
\end{array}\)
