Giải bài 93 trang 131 SGK giải tích nâng cao 12
Giải các phương trình:
a) \({{32}^{\frac{x+5}{x-7}}}=0,{{25.128}^{\frac{x+17}{x-3}}};\) b) \({{5}^{x-1}}={{10}^{x}}{{.2}^{-x}}{{.5}^{x+1}}; \)
c) \({{4}^{x}}-{{3}^{x-0,5}}={{3}^{x+0,5}}-{{2}^{2x-1}};\) d) \({{3}^{4x+8}}-{{4.3}^{2x+5}}+28=2{{\log }_{2}}\sqrt{2}\).
a) \({{32}^{\frac{x+5}{x-7}}}=0,{{25.128}^{\frac{x+17}{x-3}}}\)
\(\Leftrightarrow {{2}^{5\frac{x+5}{x-7}}}={{2}^{-2}}{{.2}^{7\frac{x+17}{x-3}}} \\ \Leftrightarrow \dfrac{5x+25}{x-7}=\dfrac{7x+119}{x-3}-2 \\ \Leftrightarrow \left( 5x+25 \right)\left( x-3 \right)=\left( 7x+119 \right)\left( x-7 \right)-2\left( x-7 \right)\left( x-3 \right) \\ \Leftrightarrow 80x=800 \\ \Leftrightarrow x=10 \)
b) \({{5}^{x-1}}={{10}^{x}}{{.2}^{-x}}{{.5}^{x+1}}\)
\( \Leftrightarrow \dfrac{{{5}^{x-1}}}{{{5}^{x}}{{.5}^{x+1}}}={{2}^{x}}{{.2}^{-x}} \\ \Leftrightarrow {{5}^{x-1-x-1-x}}={{2}^{x-x}} \\ \Leftrightarrow {{5}^{-x-2}}=1 \\ \Leftrightarrow x+2=0 \\ \Leftrightarrow x=-2 \)
c) \({{4}^{x}}-{{3}^{x-0,5}}={{3}^{x+0,5}}-{{2}^{2x-1}}\)
\(\Leftrightarrow {{4}^{x}}+\dfrac{1}{2}{{.4}^{x}}={{3}^{x-0,5}}+{{3}^{x+0,5}} \\ \Leftrightarrow \dfrac{3}{2}{{.4}^{x}}={{3}^{x-0,5}}\left( 1+3 \right) \\ \Leftrightarrow \dfrac{{{4}^{x}}}{{{4}^{0,5}}.4}=\dfrac{{{3}^{x-0,5}}}{3} \\ \Leftrightarrow {{4}^{x-1,5}}={{3}^{x-1,5}} \\ \Leftrightarrow {{\left( \dfrac{4}{3} \right)}^{x-1,5}}=1 \\ \Leftrightarrow x-1,5=0 \\ \Leftrightarrow x=1,5 \)
d) \({{3}^{4x+8}}-{{4.3}^{2x+5}}+28=2{{\log }_{2}}\sqrt{2}\)
\({{3}^{2\left( 2x+4 \right)}}-{{12.3}^{2x+4}}+28-1=0 \)
Đặt \( {{3}^{2x+4}}=t,t>0 \) phương trình trở thành.
\( {{t}^{2}}-12t+27=0 \\ \Leftrightarrow \left[ \begin{aligned} & t=9 \\ & t=3 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & {{3}^{2x+4}}=9 \\ & {{3}^{2x+4}}=3 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & 2x+4=2 \\ & 2x+4=1 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=-1 \\ & x=-\dfrac{3}{2} \\ \end{aligned} \right. \)