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