Đáp án:
d. x=5
Giải thích các bước giải:
\(\begin{array}{l}
a.\sqrt {x + 7} = x\\
\to x + 7 = {x^2}\left( {x \ge 0} \right)\\
\to \left[ \begin{array}{l}
x = \dfrac{{1 + \sqrt {29} }}{2}\\
x = \dfrac{{1 - \sqrt {29} }}{2}\left( l \right)
\end{array} \right.\\
b.\sqrt {x + 2} = x + 1\\
\to x + 2 = {x^2} + 2x + 1\left( {DK:x \ge - 1} \right)\\
\to {x^2} + x - 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 1 - \sqrt 5 }}{2}\left( l \right)
\end{array} \right.\\
c.\sqrt {{x^2} + 3x + 3} = 2x + 1\\
\to {x^2} + 3x + 3 = 4{x^2} + 4x + 1\left( {DK:x \ge - \dfrac{1}{2}} \right)\\
\to 3{x^2} + x - 2 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{2}{3}\\
x = - 1\left( l \right)
\end{array} \right.\\
d.\sqrt {2x - 1} + 2 = x\\
\to \sqrt {2x - 1} = x - 2\\
\to 2x - 1 = {x^2} - 4x + 4\left( {x \ge 2} \right)\\
\to {x^2} - 6x + 5 = 0\\
\to \left[ \begin{array}{l}
x = 5\\
x = 1\left( l \right)
\end{array} \right.\\
e.\sqrt {{x^2} + 3x} = 2x + 4\\
\to {x^2} + 3x = 4{x^2} + 16x + 16\left( {x \ge 0} \right)\\
\to 3{x^2} + 13x + 16 = 0\\
\to x \in \emptyset \\
f.\sqrt {x + 2} = x + 1\\
\to x + 2 = {x^2} + 2x + 1\left( {DK:x \ge - 1} \right)\\
\to {x^2} + x - 1 = 0\\
\to \left[ \begin{array}{l}
x = \dfrac{{ - 1 + \sqrt 5 }}{2}\\
x = \dfrac{{ - 1 - \sqrt 5 }}{2}\left( l \right)
\end{array} \right.
\end{array}\)
