Câu 29 trang 121 SGK Đại số 10 nâng cao, Giải các hệ bất phương trình...

Giải các hệ bất phương trình

a)

$\left\{ \matrix{ {{5x + 2} \over 3} \ge 4 - x \hfill \cr {{6 - 5x} \over {13}} < 3x + 1 \hfill \cr} \right.$

b)

$\left\{ \matrix{ {(1 - x)^2} > 5 + 3x + {x^2} \hfill \cr {(x + 2)^3} < {x^3} + 6{x^2} - 7x - 5 \hfill \cr} \right.$

c)

$\left\{ \matrix{ {{4x - 5} \over 7}< x + 3 \hfill \cr {{3x + 8} \over 4} > 2x - 5 \hfill \cr} \right.$

d)

$\left\{ \matrix{ x - 1 \le 2x - 3 \hfill \cr 3x < x + 5 \hfill \cr {{5 - 3x} \over 2} \le x - 3 \hfill \cr} \right.$

Đáp án

a) Ta có:

$\eqalign{ & \left\{ \matrix{ {{5x + 2} \over 3} \ge 4 - x \hfill \cr {{6 - 5x} \over {13}} < 3x + 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 5x + 2 \ge 12 - 3x \hfill \cr 6 - 5x < 39x + 13 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ 8x \ge 10 \hfill \cr 44x > - 7 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x \ge {5 \over 4} \hfill \cr x > - {7 \over {44}} \hfill \cr} \right. \Leftrightarrow x \ge {5 \over 4} \cr}$

Vậy $S = {\rm{[}}{5 \over 4}; + \infty )$

b) Ta có:

$\eqalign{ & \left\{ \matrix{ {(1 - x)^2} > 5 + 3x + {x^2} \hfill \cr {(x + 2)^3} < {x^3} + 6{x^2} - 7x - 5 \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{ 1 - 2x + {x^2} > 5 + 3x + {x^2} \hfill \cr {x^3} + 6{x^2} + 12x + 8 < {x^3} + 6{x^2} - 7x - 5 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ 5x < - 4 \hfill \cr 19x < - 13 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x < - {4 \over 5} \hfill \cr x < - {{13} \over {19}} \hfill \cr} \right. \Leftrightarrow x < - {4 \over 5} \cr}$

Vậy $S = ( - \infty ; - {4 \over 5})$

c) Ta có:

$\eqalign{ & \left\{ \matrix{ {{4x - 5} \over 7} < x + 3 \hfill \cr {{3x + 8} \over 4} > 2x - 5 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ 4x - 5 < 7x + 21 \hfill \cr 3x + 8 > 8x - 20 \hfill \cr} \right. \cr&\Leftrightarrow \left\{ \matrix{ 3x > - 26 \hfill \cr 5x < 28 \hfill \cr} \right. \cr & \Leftrightarrow \left\{ \matrix{ x > - {{26} \over 3} \hfill \cr x < {{28} \over 5} \hfill \cr} \right. \Leftrightarrow - {{26} \over 3} < x < {{28} \over 5} \cr}$

Vậy $S = ( - {{26} \over 3};{{28} \over 5})$

d) Ta có:

$\left\{ \matrix{ x - 1 \le 2x - 3 \hfill \cr 3x < x + 5 \hfill \cr {{5 - 3x} \over 2} \le x - 3 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{ x \ge 2 \hfill \cr 2x < 5 \hfill \cr 5 - 3x \le 2x - 6 \hfill \cr} \right.$

$\Leftrightarrow \left\{ \matrix{ x \ge 2 \hfill \cr x < {5 \over 2} \hfill \cr 5x \ge 11 \hfill \cr} \right.\Leftrightarrow {{11} \over 5} \le x <{5 \over 2}$

Vậy $S = {\rm{[}}{{11} \over 5};{5 \over 2})$