Giải bài 69 trang 124 SGK giải tích nâng cao 12

a) Điều kiện: $x>0$

${{\log }^{2}}{{x}^{3}}-20\log \sqrt{x}+1=0 \\ \Leftrightarrow {{\left( 3\log x \right)}^{2}}-20.\dfrac{1}{2}\log x+1=0 \\ \Leftrightarrow 9{{\log }^{2}}x-10\,\log x+1=0 \\ \Leftrightarrow \left[ \begin{aligned} & \log x=1 \\ & \log x=\dfrac{1}{9} \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x=10 \\ & x={{10}^{\frac{1}{9}}} \\ \end{aligned} \right.$

b) Điều kiện: $x>0,x\ne 1$

$\dfrac{{{\log }_{2}}x}{{{\log }_{4}}2x}=\dfrac{{{\log }_{8}}4x}{{{\log }_{16}}8x} \\ \Leftrightarrow {{\log }_{2}}x{{\log }_{16}}8x={{\log }_{8}}4x.{{\log }_{4}}2x$

Ta có

${{\log }_{16}}8x=\dfrac{1}{4}{{\log }_{2}}8x=\dfrac{3+{{\log }_{2}}x}{4} \\ {{\log }_{8}}4x=\dfrac{1}{3}{{\log }_{2}}4x=\dfrac{2+{{\log }_{2}}x}{3} \\ {{\log }_{4}}2x=\dfrac{1}{2}{{\log }_{2}}2x=\dfrac{1+{{\log }_{2}}x}{2}$

Đặt ${{\log }_{2}}x=t$ phương trình trở thành

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

c) Điều kiện: $x>0,x\ne 1$.

${{\log }_{9x}}27-{{\log }_{3x}}3+{{\log }_{9}}243=0 \\ \Leftrightarrow 3{{\log }_{9x}}3-{{\log }_{3x}}3+{{\log }_{{{3}^{2}}}}{{3}^{5}}=0 \\ \Leftrightarrow \dfrac{3}{{{\log }_{3}}9x}-\dfrac{1}{{{\log }_{3}}3x}+\dfrac{5}{2}{{\log }_{3}}3=0 \\ \Leftrightarrow \dfrac{3}{2+{{\log }_{3}}x}-\dfrac{1}{1+{{\log }_{3}}x}+\dfrac{5}{2}=0$

Đặt ${{\log }_{3}}x=t$ phương trình trở thành

$\dfrac{3}{2+t}-\dfrac{1}{1+t}+\dfrac{5}{2}=0 \\ \Leftrightarrow 6+6t-4-2t+5\left( {{t}^{2}}+3t+2 \right)=0 \\ \Leftrightarrow 5{{t}^{2}}+19t+12=0 \\ \Leftrightarrow \left[ \begin{aligned} & t=-3 \\ & t=-0,8 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & {{\log }_{3}}x=-3 \\ & {{\log }_{3}}x=-0,8 \\ \end{aligned} \right. \\ \Leftrightarrow \left[ \begin{aligned} & x={{3}^{-3}} \\ & x={{3}^{-0,8}} \\ \end{aligned} \right.$