Bài 67 trang 151 SGK Đại số 10 nâng cao


    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 ána) Ta có:\(\eqalign{ & \sqrt {{x^2} + x -...

    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 \(S = (-∞, -2]\)