Giải bài 94 trang 131 SGK giải tích nâng cao 12

a) Điều kiện: $x>0$

${{\log }_{3}}\left( \log _{0,5}^{2}x-3{{\log }_{0,5}}x+5 \right)=2 \\ \Leftrightarrow \log _{0,5}^{2}x-3{{\log }_{0,5}}x+5={{3}^{2}} \\ \Leftrightarrow \log _{0,5}^{2}x-3{{\log }_{0,5}}x-4=0 \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{0,5}}x=-1 \\ & {{\log }_{0,5}}x=4 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=2 \\ & x=\dfrac{1}{16} \\ \end{aligned} \right.$

b) Điều kiện: $\left\{ \begin{aligned} & {{4.3}^{x}}-6>0 \\ & {{9}^{x}}-6>0 \\ \end{aligned} \right.$

${{\log }_{2}}\left( {{4.3}^{x}}-6 \right)-{{\log }_{2}}\left( {{9}^{x}}-6 \right)=1 \\ \Leftrightarrow {{\log }_{2}}\left( {{4.3}^{x}}-6 \right)={{\log }_{2}}\left[ 2\left( {{9}^{x}}-6 \right) \right] \\ \Leftrightarrow {{4.3}^{x}}-6={{2.9}^{x}}-12 \\ \Leftrightarrow {{2.3}^{2x}}-{{4.3}^{x}}-6=0 \\ \Leftrightarrow \left[ \begin{aligned} & {{3}^{x}}=-1\,\left( vn \right) \\ & {{3}^{x}}=3 \\ \end{aligned} \right. \\ \Leftrightarrow x=1\left( \text{thỏa mãn} \right)$

c) Điều kiện: $x>9$

$1-\dfrac{1}{2}\log \left( 2x-1 \right)=\dfrac{1}{2}\log \left( x-9 \right) \\ \Leftrightarrow 2=\log \left( 2x-1 \right)+\log \left( x-9 \right) \\ \Leftrightarrow \left( 2x-1 \right)\left( x-9 \right)={{10}^{2}} \\ \Leftrightarrow 2{{x}^{2}}-19x-91=0 \\ \Leftrightarrow \left[ \begin{aligned} & x=13 \\ & x=-3,5\,\left( \text{loại} \right) \\ \end{aligned} \right.$

d) Điều kiện: $x>2$

$\dfrac{1}{6}{{\log }_{2}}\left( x-2 \right)-\dfrac{1}{3}={{\log }_{\frac{1}{8}}}\sqrt{3x-5} \\ \Leftrightarrow\dfrac{1}{6}{{\log }_{2}}\left( x-2 \right)+\dfrac{1}{6}{{\log }_{2}}\left( 3x-5 \right)=\dfrac{1}{3} \\ \Leftrightarrow {{\log }_{2}}\left( x-2 \right)+{{\log }_{2}}\left( 3x-5 \right)=2 \\ \Leftrightarrow \left( x-2 \right)\left( 3x-5 \right)={{2}^{2}} \\ \Leftrightarrow 3{{x}^{2}}-11x+6=0 \\ \Leftrightarrow \left[ \begin{aligned} & x=3 \\ & x=\dfrac{2}{3}\,\left( \text{loại} \right) \\ \end{aligned} \right.$