Câu 29 trang 121 SGK Đại số 10 nâng cao

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3.7 trên 3 phiếu

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

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