Giải bài 2.70 trang 133 - SBT Giải tích lớp 12
Giải các bất phương trình mũ sau :
\(\begin{align} & a)\,{{\left( 8,4 \right)}^{\frac{x-3}{{{x}^{2}}+1}}}<1; \\ & b)\,{{2}^{\left| x-2 \right|}}>{{4}^{\left| x+1 \right|}}; \\ & c)\,\dfrac{{{4}^{x}}-{{2}^{x+1}}+8}{{{2}^{1-x}}}<{{8}^{x}}; \\ & d)\,\dfrac{1}{{{3}^{x}}+5}\le \dfrac{1}{{{3}^{x+1}}-1}. \\ \end{align} \)
a)
\(\begin{aligned} & {{\left( 8,4 \right)}^{\frac{x-3}{{{x}^{2}}+1}}}<1 \\ & \Leftrightarrow \dfrac{x-3}{{{x}^{2}}+1}<{{\log }_{8,4}}1=0 \\ & \Leftrightarrow x-3<0 \\ & \Leftrightarrow x<3 \\ \end{aligned} \)
b)
\(\begin{aligned} & {{2}^{\left| x-2 \right|}}>{{4}^{\left| x+1 \right|}} \\ & \Leftrightarrow {{2}^{\left| x-2 \right|}}>{{2}^{2\left| x+1 \right|}} \\ & \Leftrightarrow \left| x-2 \right|>2\left| x+1 \right| \\ & \Leftrightarrow {{x}^{2}}-4x+4>4{{x}^{2}}+8x+4 \\ & \Leftrightarrow 3{{x}^{2}}+12x<0 \\ & \Leftrightarrow -4< x<0 \\ \end{aligned} \)
c)
\(\begin{aligned} & \dfrac{{{4}^{x}}-{{2}^{x+1}}+8}{{{2}^{1-x}}}<{{8}^{x}} \\ & \Leftrightarrow {{4}^{x}}-{{2}^{x+1}}+8<{{2}^{1-x}}{{.2}^{3x}} \\ & \Leftrightarrow {{2}^{2x}}-{{2.2}^{x}}+8<{{2}^{2x+1}} \\ & \Leftrightarrow -{{2}^{2x}}-{{2.2}^{x}}+8<0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{2}^{x}}<-4\,\,\,\,\left( \text{vô lí} \right) \\ & {{2}^{x}}>2 \\ \end{aligned} \right.\Leftrightarrow {{2}^{-x}}>2 \Leftrightarrow x>1 \\ \end{aligned} \)
d)
\(\begin{aligned} & \dfrac{1}{{{3}^{x}}+5}\le \dfrac{1}{{{3}^{x+1}}-1} \\ & \Leftrightarrow \dfrac{1}{{{3}^{x}}+5}-\dfrac{1}{{{3}^{x+1}}-1}\le 0 \\ & \Leftrightarrow \dfrac{{{3.3}^{x}}-1-\left( {{3}^{x}}+5 \right)}{\left( {{3}^{x}}+5 \right)\left( {{3.3}^{x}}-1 \right)}\le 0 \\ & \Leftrightarrow \dfrac{{{2.3}^{x}}-6}{\left( {{3}^{x}}+5 \right)\left( {{3.3}^{x}}-1 \right)}\le 0 \\ & \Leftrightarrow \dfrac{1}{3}<{{3}^{x}}\le 3 \\ & \Leftrightarrow -1< x\le 1 \\ \end{aligned} \)