# Bài 73 trang 154 SGK Đại số 10 nâng cao

Bình chọn:
3.6 trên 5 phiếu

## Giải các bất phương trình sau

Giải các bất phương trình sau

a) $$\sqrt {{x^2} - x - 12} \ge x - 1$$

b) $$\sqrt {{x^2} - 4x - 12} > 2x + 3$$

c) $${{\sqrt {x + 5} } \over {1 - x}} < 1$$

Đáp án

a) Ta có:

\eqalign{ & \sqrt {{x^2} - x - 12} \ge x - 1\cr& \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ x - 1 < 0 \hfill \cr {x^2} - x - 12 \ge 0 \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x - 1 \ge 0 \hfill \cr {x^2} - x - 12 \ge {(x - 1)^2} \hfill \cr} \right. \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ x < 1 \hfill \cr \left[ \matrix{ x \le - 3 \hfill \cr x \ge 4 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x \ge 1 \hfill \cr x \ge 13 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x \le - 3 \hfill \cr x \ge 13 \hfill \cr} \right. \cr}

Vậy $$S = (-∞, -3] ∪ [13, +∞)$$

b) Ta có:

\eqalign{ & \sqrt {{x^2} - 4x - 12} > 2x + 3 \cr&\Leftrightarrow \left[ \matrix{ \left\{ \matrix{ 2x + 3 < 0 \hfill \cr {x^2} - 4x - 12 \ge 0 \hfill \cr} \right. \hfill \cr \left\{ \matrix{ 2x - 3 \ge 0 \hfill \cr {x^2} - 4x - 12 > {(2x + 3)^2} \hfill \cr} \right. \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ x < - {3 \over 2} \hfill \cr \left[ \matrix{ x \le - 2 \hfill \cr x \ge 6 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x \ge {3 \over 2} \hfill \cr 3{x^2} + 16x + 21 < 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{ x \le - 2 \hfill \cr \left\{ \matrix{ x \ge {3 \over 2} \hfill \cr - 3 < x < - {7 \over 3} \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow x < - 2 \cr}

Vậy $$S = (-∞, -2]$$

c) Bất phương trình đã cho tương đương với:

$$(I)\,\left\{ \matrix{ 1 - x > 0 \hfill \cr \sqrt {x + 5} < 1 - x \hfill \cr} \right.\,\,\,\,;\,\,\,\,(II)\left\{ \matrix{ 1 - x < 0 \hfill \cr \sqrt {x + 5} > 1 - x \hfill \cr} \right.$$

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

Vậy $$S = [-5, -1) ∪ (1, +∞)$$

Loigiaihay.com

Các bài liên quan

Các bài khác cùng chuyên mục