Giải bài 2.85; 2.86; 2.87 trang 135 - SBT Giải tích lớp 12

2.85.

Điều kiện: \(x>1\)

\(\begin{aligned} & {{\log }_{2}}x+{{\log }_{2}}\left( x-1 \right)=1 \\ & \Leftrightarrow {{\log }_{2}}\left[ x\left( x-1 \right) \right]=1 \\ & \Leftrightarrow {{x}^{2}}-x=2 \\ & \Leftrightarrow {{x}^{2}}-x-2=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=-1\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \text{loại} \right) \\ & x=2 \\ \end{aligned} \right. \\ \end{aligned} \)

Chọn B.

2.86. ĐK:  \( x-3>0;{{x}^{2}}-6x+7>0 \)

\(\begin{aligned} & \lg \left( {{x}^{2}}-6x+7 \right)=\lg \left( x-3 \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ & \Rightarrow {{x}^{2}}-6x+7=x-3 \\ & \Leftrightarrow {{x}^{2}}-7x+10=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=2\,\text{(loại)} \\ & x=5 \,\text{(thỏa mãn)}\\ \end{aligned} \right. \\ \end{aligned}\)

Chọn B

2.87.

\(\begin{aligned} & \dfrac{{{\log }_{2}}x}{{{\log }_{4}}2x}=\dfrac{{{\log }_{8}}4x}{{{\log }_{16}}8x}\,\,\,\,\,\,\,\,\left( x>0 \right) \\ & \Rightarrow {{\log }_{2}}x.{{\log }_{16}}8x={{\log }_{8}}4x.{{\log }_{4}}2x \\ & \Leftrightarrow {{\log }_{2}}x.\left[ \dfrac{1}{4}.\left( 3+{{\log }_{2}}x \right) \right]=\dfrac{1}{3}.\left( 2+{{\log }_{2}}x \right).\left[ \dfrac{1}{2}.\left( 1+{{\log }_{2}}x \right) \right] \\ & \Leftrightarrow 3{{\log }_{2}}x\left( 3+{{\log }_{2}}x \right)=2\left( 2+{{\log }_{2}}x \right)\left( 1+{{\log }_{2}}x \right) \\ & \Leftrightarrow 3\log _{2}^{2}x+9{{\log }_{2}}x=2\log _{2}^{2}x+6{{\log }_{2}}x+4 \\ & \Leftrightarrow \log _{2}^{2}x+3{{\log }_{2}}x-4=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{\log }_{2}}x=1 \\ & {{\log }_{2}}x=-4 \\ \end{aligned} \right.\Leftrightarrow \left[ \begin{aligned} & x=2 \\ & x=\dfrac{1}{16} \\ \end{aligned} \right. \\ \end{aligned} \)

Chọn D.