Giải bài 2.46 trang 124 - SBT Giải tích lớp 12

Hướng dẫn:

Biến đổi phương trình về dạng:

 \(a^{f(x)}=a^{g(x)}\\ \Rightarrow f(x)=g(x)\)

a)
 
\(\begin{aligned} & {{\left( 0,75 \right)}^{2x-3}}={{\left( 1\dfrac{1}{3} \right)}^{5-x}} \\ & \Leftrightarrow {{\left( \dfrac{3}{4} \right)}^{2x-3}}={{\left( \dfrac{4}{3} \right)}^{5-x}}={{\left( \dfrac{3}{4} \right)}^{x-5}} \\ & \Leftrightarrow 2x-3=x-5 \\ & \Leftrightarrow x=-2 \\ \end{aligned}\)

b)

\(\begin{aligned} & {{5}^{{{x}^{2}}-5x-6}}=1 \\ & \Leftrightarrow {{x}^{2}}-5x-6=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=-1 \\ & x=6 \\ \end{aligned} \right. \\ \end{aligned} \)

c)

\(\begin{aligned} & {{\left( \dfrac{1}{7} \right)}^{{{x}^{2}}-2x-3}}={{7}^{x+1}} \\ & \Leftrightarrow {{7}^{-{{x}^{2}}+2x+3}}={{7}^{x+1}} \\ & \Leftrightarrow -{{x}^{2}}+2x+3=x+1 \\ & \Leftrightarrow -{{x}^{2}}+x+2=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=-1 \\ & x=2 \\ \end{aligned} \right. \\ \end{aligned}\)

d) ĐKXĐ: \(x\ne 7;x\ne 3 \)

\(\begin{aligned} & {{32}^{\frac{x+5}{x-7}}}=0,{{25.125}^{\frac{x+17}{x-3}}} \\ & \Leftrightarrow {{2}^{\frac{5\left( x+5 \right)}{x-7}}}=\frac{1}{4}{{.5}^{\frac{3\left( x+17 \right)}{x-3}}} \\ & \Leftrightarrow {{2}^{\frac{5\left( x+5 \right)}{x-7}+2}}={{5}^{\frac{3x+51}{x-3}}} \\ & \Leftrightarrow \dfrac{5x+25+2x-14}{x-7}=\dfrac{3x+51}{x-3}.{{\log }_{2}}5 \\ & \Leftrightarrow \dfrac{7x+11}{x-7}=\dfrac{3x+51}{x-3}.{{\log }_{2}}5 \\ & \Rightarrow \left( 7x+11 \right)\left( x-3 \right)=\left( 3x+51 \right)\left( x-7 \right).{{\log }_{2}}5 \\ & \Leftrightarrow \left( 7-3{{\log }_{2}}5 \right).{{x}^{2}}-2\left( 5+15{{\log }_{2}}5 \right)x-\left( 33-357{{\log }_{2}}5 \right)=0 \\ \end{aligned} \)

Ta có:

\(\Delta '=1296\log _{2}^{2}5-2448{{\log }_{2}}5+256>0 \)

Phương trình có hai nghiệm

\(x=\dfrac{5+15{{\log }_{2}}5\pm \sqrt{\Delta '}}{7-3{{\log }_{2}}5} \) (TMĐK)