Giải bài 2.47 trang 124 - SBT Giải tích lớp 12
Giải các phương trình mũ sau:
\(\begin{align} & a)\,{{2}^{x+4}}+{{2}^{x+2}}={{5}^{x+1}}+{{3.5}^{x}} \\ & b)\,{{5}^{2x}}-{{7}^{x}}-{{5}^{2x}}.17+{{7}^{x}}.17=0 \\ & c)\,{{4.9}^{x}}+12^x-{{3.16}^{x}}=0 \\ & d)\,-{{8}^{x}}+{{2.4}^{x}}+{{2}^{x}}-2=0 \\ \end{align} \)
Gợi ý:
a) Chia cả hai vế cho \(5^x\)
b) Rút gọn vế trái
c) Chia hai vế cho \(12^x\)
d) Đưa về phương trình bậc ba ẩn \(2^x\)
a)
\(\begin{aligned} & {{2}^{x+4}}+{{2}^{x+2}}={{5}^{x+1}}+{{3.5}^{x}} \\ & \Leftrightarrow 16.{{\left( \dfrac{2}{5} \right)}^{x}}+4.{{\left( \dfrac{2}{5} \right)}^{x}}=5+3 \\ & \Leftrightarrow 20.{{\left( \dfrac{2}{5} \right)}^{x}}=8 \\ & \Leftrightarrow {{\left( \dfrac{2}{5} \right)}^{x}}=\dfrac{2}{5} \\ & \Leftrightarrow x=1 \\ \end{aligned} \)
b)
\(\begin{aligned} & {{5}^{2x}}-{{7}^{x}}-{{5}^{2x}}.17+{{7}^{x}}.17=0 \\ & \Leftrightarrow {{16.7}^{x}}-{{16.5}^{x}}=0 \\ & \Leftrightarrow {{7}^{x}}={{5}^{x}} \\ & \Leftrightarrow x=0 \\ \end{aligned} \)
c)
\(\begin{aligned} & {{4.9}^{x}}+{{12}^{x}}-{{3.16}^{x}}=0 \\ & \Leftrightarrow 4.{{\left( \dfrac{9}{12} \right)}^{x}}-3.{{\left( \dfrac{16}{12} \right)}^{x}}+1=0 \\ & \Leftrightarrow 4.{{\left( \dfrac{3}{4} \right)}^{x}}-3.{{\left( \dfrac{4}{3} \right)}^{x}}+1=0 \\ & \Leftrightarrow 4.{{\left( \dfrac{3}{4} \right)}^{x}}-3.\dfrac{1}{{{\left( \dfrac{3}{4} \right)}^{x}}}+1=0 \\ & \Leftrightarrow 4.{{\left( \dfrac{3}{4} \right)}^{2x}}+{{\left( \dfrac{3}{4} \right)}^{x}}-3=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\left( \dfrac{3}{4} \right)}^{x}}=-1\,\,\,\,\,\left( \text{loại} \right) \\ & {{\left( \dfrac{3}{4} \right)}^{x}}=\dfrac{3}{4} \\ \end{aligned} \right. \\ & \Leftrightarrow x=1 \\ \end{aligned} \)
d)
\(\begin{aligned} & -{{8}^{x}}+{{2.4}^{x}}+{{2}^{x}}-2=0 \\ & \Leftrightarrow -{{2}^{3x}}+{{2.2}^{2x}}+{{2}^{x}}-2=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{2}^{x}}=1 \\ & {{2}^{x}}=-1\,\,\,\,\left( \text{loại} \right) \\ & {{2}^{x}}=2 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=0 \\ & x=1 \\ \end{aligned} \right. \\ \end{aligned} \)