Bài 62 trang 136 SGK Đại số 10 nâng cao


    Giải các hệ bất phương trìnha) \(\left\{ \matrix{ 4x - 3 < 3x + 4 \hfill \cr {x^2} - 7x + 10 \le 0 \hfill \cr} \right.\)b) \(\left\{ \matrix{ 2{x^2} + 9x - 7 > 0 \hfill \cr {x^2} + x - 6 \le 0 \hfill \cr} \right.\)c)\(\left\{ \matrix{ {x^2} - 9...

    Giải các hệ bất phương trình

    a) 

    \(\left\{ \matrix{
    4x – 3 < 3x + 4 \hfill \cr
    {x^2} – 7x + 10 \le 0 \hfill \cr} \right.\)

    b) 

    \(\left\{ \matrix{
    2{x^2} + 9x – 7 > 0 \hfill \cr
    {x^2} + x – 6 \le 0 \hfill \cr} \right.\)

    c)

    \(\left\{ \matrix{
    {x^2} – 9 < 0 \hfill \cr
    (x – 1)(3{x^2} + 7x + 4) \ge 0 \hfill \cr} \right.\)

    Đáp án

    a) Ta có: 

    \( \Leftrightarrow \left\{ \matrix{
    4x – 3 < 3x + 4 \hfill \cr
    {x^2} – 7x + 10 \le 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
    x < 7 \hfill \cr
    2 \le x \le 5 \hfill \cr} \right.\)

    \(\Leftrightarrow 2 \le x \le 5\)

    Vậy \(S = [2, 5]\)

    b) Ta có:

    \(\eqalign{
    & \left\{ \matrix{
    2{x^2} + 9x – 7 > 0 \hfill \cr
    {x^2} + x – 6 \le 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
    \left[ \matrix{
    x < {{ – 9 – \sqrt {137} } \over 4} \hfill \cr
    x > {{ – 9 + \sqrt {137} } \over 4} \hfill \cr} \right. \hfill \cr
    – 3 \le x \le 2 \hfill \cr} \right. \cr
    & \Leftrightarrow {{ – 9 + \sqrt {137} } \over 4} < x < 2 \cr} \) 

    Vậy \(S = ({{ – 9 + \sqrt {137} } \over 4};2{\rm{]}}\)

    c) Ta có:

    \(\eqalign{
    & \left\{ \matrix{
    {x^2} – 9 < 0 \hfill \cr
    (x – 1)(3{x^2} + 7x + 4) \ge 0 \hfill \cr} \right.\cr& \Leftrightarrow \left\{ \matrix{
    – 3 < x < 3 \hfill \cr
    \left[ \matrix{
    – {4 \over 3} \le x \le – 1 \hfill \cr
    x \ge 1 \hfill \cr} \right. \hfill \cr} \right. \cr
    & \Leftrightarrow \left[ \matrix{
    – {4 \over 3} \le x \le – 1 \hfill \cr
    1 \le x \le 3 \hfill \cr} \right. \cr} \)

    Vậy \(S = \,{\rm{[}} – {4 \over 3},\, – 1{\rm{]}}\, \cup {\rm{[}}1,\,3)\)