Bài 73 trang 154 SGK Đại số 10 nâng cao


    Giải các bất phương trình saua) \(\sqrt {{x^2} - x - 12}  \ge x - 1\)b) \(\sqrt {{x^2} - 4x - 12}  > 2x + 3\)c) \({{\sqrt {x + 5} } \over {1 - x}} < 1\)Đáp ána) Ta có:\(\eqalign{ & \sqrt {{x^2} - x - 12} \ge x -...

    Giải các bất phương trình sau

    a) \(\sqrt {{x^2} – x – 12}  \ge x – 1\)

    b) \(\sqrt {{x^2} – 4x – 12}  > 2x + 3\)

    c) \({{\sqrt {x + 5} } \over {1 – x}} < 1\)

    Đáp án

    a) Ta có:

    \(\eqalign{
    & \sqrt {{x^2} – x – 12} \ge x – 1\cr& \Leftrightarrow \left[ \matrix{
    \left\{ \matrix{
    x – 1 < 0 \hfill \cr
    {x^2} – x – 12 \ge 0 \hfill \cr} \right. \hfill \cr
    \left\{ \matrix{
    x – 1 \ge 0 \hfill \cr
    {x^2} – x – 12 \ge {(x – 1)^2} \hfill \cr} \right. \hfill \cr} \right. \cr
    & \Leftrightarrow \left[ \matrix{
    \left\{ \matrix{
    x < 1 \hfill \cr
    \left[ \matrix{
    x \le – 3 \hfill \cr
    x \ge 4 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr
    \left\{ \matrix{
    x \ge 1 \hfill \cr
    x \ge 13 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
    x \le – 3 \hfill \cr
    x \ge 13 \hfill \cr} \right. \cr} \) 

    Vậy \(S = (-∞, -3] ∪ [13, +∞)\)

    b) Ta có:

    \(\eqalign{
    & \sqrt {{x^2} – 4x – 12} > 2x + 3 \cr&\Leftrightarrow \left[ \matrix{
    \left\{ \matrix{
    2x + 3 < 0 \hfill \cr
    {x^2} – 4x – 12 \ge 0 \hfill \cr} \right. \hfill \cr
    \left\{ \matrix{
    2x – 3 \ge 0 \hfill \cr
    {x^2} – 4x – 12 > {(2x + 3)^2} \hfill \cr} \right. \hfill \cr} \right. \cr
    & \Leftrightarrow \left[ \matrix{
    \left\{ \matrix{
    x < – {3 \over 2} \hfill \cr
    \left[ \matrix{
    x \le – 2 \hfill \cr
    x \ge 6 \hfill \cr} \right. \hfill \cr} \right. \hfill \cr
    \left\{ \matrix{
    x \ge {3 \over 2} \hfill \cr
    3{x^2} + 16x + 21 < 0 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
    x \le – 2 \hfill \cr
    \left\{ \matrix{
    x \ge {3 \over 2} \hfill \cr
    – 3 < x < – {7 \over 3} \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow x < – 2 \cr} \)

    Vậy \(S = (-∞, -2]\)

    c) Bất phương trình đã cho tương đương với:

    \((I)\,\left\{ \matrix{
    1 – x > 0 \hfill \cr
    \sqrt {x + 5} < 1 – x \hfill \cr} \right.\,\,\,\,;\,\,\,\,(II)\left\{ \matrix{
    1 – x < 0 \hfill \cr
    \sqrt {x + 5} > 1 – x \hfill \cr} \right.\)

    \(\eqalign{
    & (I) \Leftrightarrow \left\{ \matrix{
    x < 1 \hfill \cr
    x + 5 \ge 0 \hfill \cr
    x + 5 < {(1 – x)^2} \hfill \cr
    – 5 \le x < 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
    x < 1 \hfill \cr
    x \ge – 5 \hfill \cr
    {x^2} – 3x – 4 > 0 \hfill \cr} \right. \cr
    & \Leftrightarrow \left\{ \matrix{
    x < 1 \hfill \cr
    x \ge – 5 \hfill \cr
    {x^2} – 3x – 4 > 0 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
    – 5 \le x < 1 \hfill \cr
    \left[ \matrix{
    x < – 1 \hfill \cr
    x > 4 \hfill \cr} \right. \hfill \cr} \right. \cr&\Leftrightarrow – 5 \le x < 1 \cr} \)

    Vậy \(S = [-5, -1) ∪ (1, +∞)\)