Đáp án:
\(\begin{array}{l}
1,\\
\left[ \begin{array}{l}
x = \dfrac{{\sqrt 5 - 3}}{2}\\
x = \dfrac{{ - \sqrt 5 - 3}}{2}
\end{array} \right.\\
5,\\
x = 0\\
2,\\
x = 1
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
DKXD:\,\,\,{x^2} + 3x + 2 \ge 0\\
\sqrt {{x^2} + 3x + 2} = 1\\
\Leftrightarrow {x^2} + 3x + 2 = {1^2}\\
\Leftrightarrow {x^2} + 3x + 2 = 1\\
\Leftrightarrow {x^2} + 3x + 1 = 0\\
\Leftrightarrow {x^2} + 3x + \dfrac{9}{4} = \dfrac{5}{4}\\
\Leftrightarrow {x^2} + 2.x.\dfrac{3}{2} + {\left( {\dfrac{3}{2}} \right)^2} = \dfrac{5}{4}\\
\Leftrightarrow {\left( {x + \dfrac{3}{2}} \right)^2} = \dfrac{5}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{3}{2} = \dfrac{{\sqrt 5 }}{2}\\
x + \dfrac{3}{2} = - \dfrac{{\sqrt 5 }}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{\sqrt 5 - 3}}{2}\\
x = \dfrac{{ - \sqrt 5 - 3}}{2}
\end{array} \right.\\
5,\\
DKXD:\,\,\,{x^2} + x + 1 \ge 0\\
\sqrt {{x^2} + x + 1} = 1 + x\\
\Leftrightarrow \left\{ \begin{array}{l}
1 + x \ge 0\\
{x^2} + x + 1 = {\left( {1 + x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
{x^2} + x + 1 = {x^2} + 2x + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x = 0
\end{array} \right.\\
\Leftrightarrow x = 0\\
2,\\
DKXD:\,\,\,x \ge 0\\
3x - 2\sqrt x = 1\\
\Leftrightarrow 3x - 2\sqrt x - 1 = 0\\
\Leftrightarrow \left( {3x - 3\sqrt x } \right) + \left( {\sqrt x - 1} \right) = 0\\
\Leftrightarrow 3\sqrt x \left( {\sqrt x - 1} \right) + \left( {\sqrt x - 1} \right) = 0\\
\Leftrightarrow \left( {\sqrt x - 1} \right)\left( {3\sqrt x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x - 1 = 0\\
3\sqrt x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 1\\
\sqrt x = - \dfrac{1}{3}
\end{array} \right.\\
\sqrt x \ge 0 \Rightarrow \sqrt x = 1 \Leftrightarrow x = 1
\end{array}\)
