Câu 8: Trang 90 - sgk giải tích 12
Giải các bất phương trình:
a) $2^{2x-1}+ 2x^{2x-2} + 2^{2x-3} \geq 448$
b) $(0,4)^{x} – (2,5)^{x+1} > 1,5$
c) $\log_{3}\left [ \log_{\frac{1}{2}}(x^{2}-1) \right ]<1$
d) $\log^{2}_{0,2}x-5\log_{0,2}x<-6$
Bài Làm:
a) $2^{2x-1}+ 2x^{2x-2} + 2^{2x-3} \geq 448$
<=> $\frac{1}{2}2^{2x}+\frac{1}{4}2^{2x}+\frac{1}{8}2^{2x}\geq 448$
<=> $\frac{7}{2}2^{2x}\geq 448$
<=> $2^{2x}\geq 512$
<=> $2^{2x}\geq 2^{9}$
<=> $2x\geq 2^{9}$
<=> $x\geq \frac{9}{2}$
b) $(0,4)^{x} – (2,5)^{x+1} > 1,5$
<=> $(\frac{2}{5})^{x}-(\frac{5}{2}).(\frac{5}{2})^{x}>1,5$
Đặt $(\frac{2}{5})^{x}=t,(t>0)$
<=> $2t^{2}-3t-5>0$
<=> $t>\frac{5}{2}$
<=> $(\frac{2}{5})^{x}>\frac{5}{2}$
<=> $x<-1$
c) $\log_{3}\left [ \log_{\frac{1}{2}}(x^{2}-1) \right ]<1$
<=> $\log_{3}\left [ \log_{\frac{1}{2}}(x^{2}-1) \right ]<\log_{3}3$
<=> $0< \log_{\frac{1}{2}}(x^{2}-1)<3$
<=> $\frac{1}{8}<x^{2}-1<1$
<=> $\frac{3}{2\sqrt{2}}<\sqrt{x}<\sqrt{2}$
d) $\log^{2}_{0,2}x-5\log_{0,2}x<-6$
Đặt $\log_{0,2}x=t(t>0)$
<=> $t^{2}-5t+6<0$
<=> $2<t<3$
<=> $2<\log_{0,2}x<3$
<=> $5^{-3}<x<5^{-2}$