# Bài 67 trang 151 SGK Đại số 10 nâng cao

Bình chọn:
3.3 trên 3 phiếu

## Giải các bất phương trình:

Giải các bất phương trình:

a) $$\sqrt {{x^2} + x - 6} < x - 1$$

b) $$\sqrt {2x - 1} \le 2x - 3$$

c) $$\sqrt {2{x^2} - 1} > 1 - x$$

d) $$\sqrt {{x^2} - 5x - 14} \ge 2x - 1$$

Đáp án

a) Ta có:

\eqalign{ & \sqrt {{x^2} + x - 6} < x - 1\cr& \Leftrightarrow \left\{ \matrix{ {x^2} + x - 6 \ge 0 \hfill \cr x - 1 > 0 \hfill \cr {x^2} + x - 6 < {(x - 1)^2} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ \left[ \matrix{ x \le 3 \hfill \cr x \ge 2 \hfill \cr} \right. \hfill \cr x > 1 \hfill \cr 3x < 7 \hfill \cr} \right. \Leftrightarrow 2 \le x < {7 \over 3} \cr}

Vậy $$S = {\rm{[}}2,{7 \over 3})$$

b) Ta có:

\eqalign{ & \sqrt {2x - 1} \le 2x - 3 \Leftrightarrow \left\{ \matrix{ 2x - 1 \ge 0 \hfill \cr 2x - 3 \ge 0 \hfill \cr 2x - 1 \le {(2x - 3)^2} \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x \ge {1 \over 2} \hfill \cr x \ge {3 \over 2} \hfill \cr 4{x^2} - 14x + 10 \ge 0 \hfill \cr} \right.\cr& \Leftrightarrow \left\{ \matrix{ x \ge {3 \over 2} \hfill \cr \left[ \matrix{ x \le 1 \hfill \cr x \ge {5 \over 2} \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \ge {5 \over 2} \cr}

Vậy $$S = {\rm{[}}{5 \over 2}; + \infty )$$

c) Ta có:

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

Vậy $$S = ( - \infty , - 1 - \sqrt 3 ) \cup ( - 1 + \sqrt 3 , + \infty )$$

d) Ta có:

\eqalign{ & \sqrt {{x^2} - 5x - 14} \ge 2x - 1 \cr & \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ 2x - 1 < 0 \hfill \cr {x^2} - 5x - 14 \ge 0 \hfill \cr} \right. \hfill \cr \left\{ \matrix{ 2x - 1 \ge 0 \hfill \cr {x^2} - 5x - 14 \ge {(2x - 1)^2} \hfill \cr} \right. \hfill \cr} \right.\cr& \Leftrightarrow \left[ \matrix{ \left\{ \matrix{ x < {1 \over 2} \hfill \cr \left[ \matrix{ x \le - 2 \hfill \cr x \ge 7 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr \left\{ \matrix{ x \ge {1 \over 2} \hfill \cr 3{x^2} + x + 15 \le 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow x \le - 2 \cr}

Vậy $$S = (-∞, -2]$$

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