Giải bài 7 trang 90 – SGK môn Giải tích lớp 12

a) \({{3}^{x+4}}+{{3.5}^{x+3}}={{5}^{x+4}}+{{3}^{x+3}};\)

\(\begin{aligned} & \Leftrightarrow {{3}^{x+4}}-{{3}^{x+3}}={{5}^{x+4}}-{{3.5}^{x+3}} \\ & \Leftrightarrow {{3.3}^{x+3}}-{{3}^{x+3}}={{5.5}^{x+3}}-{{3.5}^{x+3}} \\ & \Leftrightarrow {{2.3}^{x+3}}={{2.5}^{x+3}} \\ & \Leftrightarrow {{\left( \dfrac{3}{5} \right)}^{x+3}}=1={{\left( \dfrac{3}{5} \right)}^{0}} \\ & \Leftrightarrow x+3=0\Leftrightarrow x=-3\\ \end{aligned} \)

Vậy \(S=\left\{ -3 \right\}\)

b) \({{25}^{x}}-{{6.5}^{x}}+5=0;\)

\(\begin{aligned} & \Leftrightarrow {{5}^{2x}}-{{6.5}^{x}}+5=0 \\ & \Leftrightarrow \left( {{5}^{x}}-1 \right)\left( {{5}^{x}}-5 \right)=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{5}^{x}}=1 \\ & {{5}^{x}}=5 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=0 \\ & x=1 \\ \end{aligned} \right.\\ \end{aligned} \)

Vậy \(S=\left\{ 0,\,1 \right\}\)

c) \({{4.9}^{x}}+{{12}^{x}}-{{3.16}^{x}}=0;\)

Chia cả hai vế cho \(16^x\) ta được:

\(4.{{\left( \dfrac{3}{4} \right)}^{2x}}+{{\left( \dfrac{3}{4} \right)}^{x}}-3=0\)

Đặt \({{\left( \dfrac{3}{4} \right)}^{x}}=t>0\) phương trình trở thành:

\(\begin{aligned} & 4{{t}^{2}}+t-3=0 \\ & \Leftrightarrow \left[ \begin{aligned} & t=-1\,\left( \text{loại} \right) \\ & t=\dfrac{3}{4} \\ \end{aligned} \right. \\ \end{aligned} \)

Với \( t=\dfrac{3}{4}\Leftrightarrow {{\left( \dfrac{3}{4} \right)}^{x}}=\dfrac{3}{4}\Leftrightarrow x=1\)

Vậy \(S=\left\{ 1 \right\}\)

d) \({{\log }_{7}}\left( x-1 \right){{\log }_{7}}x={{\log }_{7}}x;\)

Điều kiện: \(x>1\)

\(\begin{aligned} &{{\log }_{7}}x\left[ {{\log }_{7}}\left( x-1 \right)-1 \right]=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\log }_{7}}x=0 \\ & {{\log }_{7}}\left( x-1 \right)=1 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x={{7}^{0}} \\ & x-1=7 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=1\,\left( \text{loại} \right) \\ & x=8 \\ \end{aligned} \right. \\ \end{aligned} \)

Vậy \(S=\left\{ 8 \right\} \)

e) \({{\log }_{3}}x+{{\log }_{\sqrt{3}}}x+{{\log }_{\frac{1}{3}}}x=6;\)

Điều kiện: \(x>0\)

\(\begin{aligned} & \Leftrightarrow {{\log }_{3}}x+2{{\log }_{3}}x-{{\log }_{3}}x=6 \\ & \Leftrightarrow {{\log }_{3}}x=3 \\ & \Leftrightarrow x={{3}^{3}}=27\\ \end{aligned} \)

Vậy \(S=\left\{ 27 \right\}\)

g) \(\log \dfrac{x+8}{x-1}=\log x.\)

Điều kiện: \(x>1\)

\(\begin{aligned} & \Leftrightarrow \dfrac{x+8}{x-1}=x \\ & \Rightarrow {{x}^{2}}-x=x+8 \\ & \Leftrightarrow {{x}^{2}}-2x-8=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=4 \\ & x=-2\,\left( \text{loại} \right) \\ \end{aligned} \right. \\ \end{aligned} \)

Vậy \(S=\{4\}\)