Đáp án:
\(\begin{array}{l}
1)x = - 1\\
2)\left[ \begin{array}{l}
2 \ge x > 1\\
x < - 10
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)DK:\left[ \begin{array}{l}
x \ge 1\\
x \le - 1
\end{array} \right.\\
\sqrt {{x^2} + x} \ge x + 1\\
\to {x^2} + x \ge {x^2} + 2x + 1\left( {DK:x \ge - 1} \right)\\
\to x \le - 1\\
KL:x = - 1\\
2)DK:\left[ \begin{array}{l}
x \ge \dfrac{{ - 5 + \sqrt {73} }}{4}\\
x \le \dfrac{{ - 5 - \sqrt {73} }}{4}
\end{array} \right.\\
\sqrt {2{x^2} + 5x - 6} > 2 - x\\
\to 2{x^2} + 5x - 6 > 4 - 4x + {x^2}\left( {DK:2 \ge x} \right)\\
\to {x^2} + 9x - 10 > 0\\
\to \left( {x + 10} \right)\left( {x - 1} \right) > 0\\
\to \left[ \begin{array}{l}
x > 1\\
x < - 10
\end{array} \right.\\
KL:\left[ \begin{array}{l}
2 \ge x > 1\\
x < - 10
\end{array} \right.
\end{array}\)
