Nội dung bài giảng
Bài 10. Giải các bất phương trình sau
a) \({{{2^x}} \over {{3^x} - {2^x}}} \le 2\)
b) \({({1 \over 2})^{{{\log }_2}({x^2} - 1)}} > 1\)
c) \({\log ^2}x + 3\log x \ge 4\)
d) \({{1 - {{\log }_4}x} \over {1 + {{\log }_2}x}} \le {1 \over 4}\)
Trả lời:
a) Ta có:
\({{{2^x}} \over {{3^x} - {2^x}}} \le 2 \Leftrightarrow {1 \over {{{({3 \over 2})}^x} - 1}} \le 2\)
Đặt \(t = {({3 \over 2})^2}(t > 0)\) , bất phương trình trở thành:
\(\eqalign{
& {1 \over {t - 1}} \le 2 \Leftrightarrow {1 \over {t - 1}} - 2 \le 0 \Leftrightarrow {{ - 2t + 3} \over {t - 1}} \le 0 \cr
& \Leftrightarrow \left[ \matrix{
0 < t < 1 \hfill \cr
t \ge {3 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
{({3 \over 2})^x} < 1 \hfill \cr
{({3 \over 2})^2} \ge {3 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x < 0 \hfill \cr
x \ge 1 \hfill \cr} \right. \cr} \)
b) Ta có:
\(\eqalign{
& {({1 \over 2})^{{{\log }_2}({x^2} - 1)}} > 1 \Leftrightarrow \left\{ \matrix{
{x^2} - 1 > 0 \hfill \cr
{\log _2}({x^2} - 1) < 0 \hfill \cr} \right. \cr
& \Leftrightarrow 0 < {x^2} - 1 < 1 \Leftrightarrow 1 < |x| < \sqrt 2 \cr
& \Leftrightarrow x \in ( - \sqrt 2 , - 1) \cup (1,\sqrt 2 ) \cr} \)
c) Điều kiện: \(x > 0\)
\(\eqalign{
& {\log ^2}x + 3\log x \ge 4 \Leftrightarrow (\log x + 4)(logx - 1) \ge 0 \cr
& \Leftrightarrow \left[ \matrix{
{\mathop{\rm logx}\nolimits} \ge 1 \hfill \cr
logx \le - 4 \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x \ge 10 \hfill \cr
0 < x \le {10^{ - 4}} \hfill \cr} \right. \cr} \)
d) Ta có:
\(\eqalign{
& {{1 - {{\log }_4}x} \over {1 + {{\log }_2}x}} \le {1 \over 4} \Leftrightarrow {{1 - {{\log }_4}x} \over {1 + 2{{\log }_4}x}} \le {1 \over 4} \cr
& \Leftrightarrow {{3 - 6{{\log }_4}x} \over {1 + 2{{\log }_4}x}}\le0 \cr
& \Leftrightarrow \left[ \matrix{
{\log _4}x \le {{ - 1} \over 2} \hfill \cr
{\log _4}x \ge {1 \over 2} \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
0 < x < {1 \over 2} \hfill \cr
x \ge 2 \hfill \cr} \right. \cr} \)