Nội dung bài giảng
Bài 84. So sánh p và q, biết:
\(\eqalign{
& a)\,{\left( {{2 \over 3}} \right)^p} > {\left( {{3 \over 2}} \right)^{ - q}} \cr
& c)\,0,{25^p} < {\left( {{1 \over 2}} \right)^{2q}} \cr} \)
\(\eqalign{
& b)\,{\left( {{8 \over 3}} \right)^{ - p}} < {\left( {{3 \over 8}} \right)^q} \cr
& d)\,{\left( {{7 \over 2}} \right)^p} < {\left( {{2 \over 7}} \right)^{p - 2q}} \cr} \)
Giải
\(\eqalign{
& a)\,{\left( {{2 \over 3}} \right)^p} > {\left( {{3 \over 2}} \right)^{ - q}} \Leftrightarrow {\left( {{2 \over 3}} \right)^p} > {\left( {{2 \over 3}} \right)^q} \Leftrightarrow p < q\,\,\left( {\text{ vì }\,\,\,{2 \over 3} < 1} \right) \cr
& b)\,{\left( {{8 \over 3}} \right)^{ - p}} < {\left( {{3 \over 8}} \right)^q} \Leftrightarrow {\left( {{3 \over 8}} \right)^p} < {\left( {{3 \over 8}} \right)^q} \Leftrightarrow p > q\,\,\left( {\text{ vì }\,\,{3 \over 8} < 1} \right) \cr
& c)\,\,0,{25^p} < {\left( {{1 \over 2}} \right)^{2q}} \Leftrightarrow {\left( {{1 \over 4}} \right)^p} < {\left( {{1 \over 4}} \right)^q} \Leftrightarrow \,\,p > q\,\,\left( {\text{ vì }\,\,{1 \over 4} < 1} \right) \cr
& d)\,\,{\left( {{7 \over 2}} \right)^p} < {\left( {{2 \over 7}} \right)^{p - 2q}} \Leftrightarrow {\left( {{7 \over 2}} \right)^p} < {\left( {{7 \over 2}} \right)^{2q - p}} \Leftrightarrow p < 2q - p\,\,\left( {\text{ vì }\,\,{7 \over 2} > 1} \right) \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow 2p < 2q \Leftrightarrow p < q \cr} \)