Bài 18 trang 223 SGK Đại số 10 Nâng cao

a) 3x² - |5x + 2| >0

b) $\sqrt {2{x^2} + 7x + 5}  > x + 1$

c) $\sqrt {{x^2} + 4x - 5}  \le x + 3$

Đáp án

a) Ta có:

$\eqalign{ & 3{x^2} - \left| {5x + 2} \right| > 0 \Leftrightarrow |5x + 2| < 3{x^2} \cr & \Leftrightarrow - 3{x^2} < 5x + 2 < 3{x^2} \cr & \Leftrightarrow \left\{ \matrix{ 3{x^2} + 5x + 2 > 0 \hfill \cr 3{x^2} - 5x - 2 > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x < - 1 \hfill \cr x > - {2 \over 3} \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x < - {1 \over 3} \hfill \cr x > 2 \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow \left[ \matrix{ x < - 1 \hfill \cr - {2 \over 3} < x < - {1 \over 3} \hfill \cr x > 2 \hfill \cr} \right. \cr}$

Vậy: $S = ( - \infty ,\, - 1) \cup ( - {2 \over 3}; - {1 \over 3}) \cup (2, + \infty )$

b) Ta có:

$\eqalign{ & \sqrt {2{x^2} + 7x + 5} > x + 1 \cr & \Leftrightarrow \,\,\left[ \matrix{ (I)\,\left\{ \matrix{ x + 1 < 0 \hfill \cr 2{x^2} + 7x + 5 \ge 0 \hfill \cr} \right. \hfill \cr (II)\left\{ \matrix{ x + 1 \ge 0 \hfill \cr 2{x^2} + 7x + 5 > {(x + 1)^2} \hfill \cr} \right.\, \hfill \cr} \right. \cr}$ 

Ta có:

$(I) \Leftrightarrow \left\{ \matrix{ x < - 1 \hfill \cr \left[ \matrix{ x \le - {5 \over 2} \hfill \cr x \ge - 1 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \le - {5 \over 2}$ 

$(II) \Leftrightarrow \left\{ \matrix{ x \ge - 1 \hfill \cr {x^2} + 5x + 4 > 0 \hfill \cr} \right.$

$\Leftrightarrow \left\{ \matrix{ x \ge - 1 \hfill \cr \left[ \matrix{ x < - 4 \hfill \cr x > - 1 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x > - 1$ 

Vậy: $S = ( - \infty ;\, - {5 \over 2}{\rm{]}}\, \cup ( - 1;\, + \infty )$

c) Ta có:

$\eqalign{ & \sqrt {{x^2} + 4x - 5} \le x + 3 \cr&\Leftrightarrow \left\{ \matrix{ x + 3 \ge 0 \hfill \cr {x^2} + 4x - 5 \ge 0 \hfill \cr {x^2} + 4x - 5 \le {(x + 3)^2} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x \ge - 3 \hfill \cr \left[ \matrix{ x \le - 5 \hfill \cr x \ge 1 \hfill \cr} \right. \hfill \cr x \ge - 7 \hfill \cr} \right. \Leftrightarrow x \ge 1 \cr}$

Vậy $S = [1, +∞)$