Bài 73 trang 154 SGK Đại số 10 nâng cao

LG a

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

Phương pháp giải:

Áp dụng 

\(\sqrt f \ge g \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
g < 0\\
f \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
g \ge 0\\
f \ge {g^2}
\end{array} \right.
\end{array} \right.\)

Lời giải chi tiết:

 Ta có:

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

\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 1 \le 0\\
{x^2} - x - 12 \ge 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 1 \ge 0\\
{x^2} - x - 12 \ge {x^2} - 2x + 1
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x \le 1\\
\left[ \begin{array}{l}
x \ge 4\\
x \le - 3
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge 1\\
x \ge 13
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le - 3\\
x \ge 13
\end{array} \right.
\end{array}\)

Vậy \(S = (-∞, -3] ∪ [13, +∞)\)

LG b

\(\sqrt {{x^2} - 4x - 12}  > 2x + 3\)

Lời giải chi tiết:

Ta có:

\(\eqalign{
& \sqrt {{x^2} - 4x - 12} > 2x + 3 \cr&\Leftrightarrow \left[ \matrix{
\left\{ \matrix{
2x + 3 < 0 \hfill \cr 
{x^2} - 4x - 12 \ge 0 \hfill \cr} \right. \hfill \cr 
\left\{ \matrix{
2x + 3 \ge 0 \hfill \cr 
{x^2} - 4x - 12 > {(2x + 3)^2} \hfill \cr} \right. \hfill \cr} \right. \cr} \)

\(\begin{array}{l}
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x < - \frac{3}{2}\\
\left[ \begin{array}{l}
x \le - 2\\
x \ge 6
\end{array} \right.
\end{array} \right.\\
\left\{ \begin{array}{l}
x \ge - \frac{3}{2}\\
{x^2} - 4x - 12 > 4{x^2} + 12x + 9
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le - 2\\
\left\{ \begin{array}{l}
x \ge - \frac{3}{2}\\
- 3{x^2} - 16x - 21 > 0
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le - 2\\
\left\{ \begin{array}{l}
x \ge - \frac{3}{2}\\
- 3 < x < - \frac{7}{3}
\end{array} \right.
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x \le - 2\\
- \frac{3}{2} \le x < - \frac{7}{3}
\end{array} \right.\\
\Leftrightarrow x \le - 2
\end{array}\)

Vậy \(S = (-∞, -2]\)

LG c

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

Phương pháp giải:

Xét các trường hợp \(1-x < 0\) và \(1-x > 0\)

Lời giải chi tiết:

Bất phương trình đã cho tương đương với:

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

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

\(\left( {II} \right) \Leftrightarrow \left\{ \begin{array}{l}
x > 1\\
x + 5 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x > 1\\
x \ge - 5
\end{array} \right.\) \(\Leftrightarrow x > 1\)

Vậy \(S = [-5, -1) ∪ (1, +∞)\)

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