Giải bài 2.50; 2.51; 2.52 trang 125 - SBT Giải tích lớp 12

2.50.

\(\begin{aligned} & {{25}^{x}}-{{6.5}^{x}}+5=0 \\ & \Leftrightarrow {{5}^{2x}}-{{6.5}^{x}}+5=0 \\ & \Leftrightarrow \left[ \begin{aligned} & {{5}^{x}}=1 \\ & {{5}^{x}}=5 \\ \end{aligned} \right. \\ & \Leftrightarrow \left[ \begin{aligned} & x=0 \\ & x=1 \\ \end{aligned} \right. \\ \end{aligned} \)

Chọn B.

2.51

\(\begin{aligned} & {{25}^{x}}-{{2.10}^{x}}+{{4}^{x}}=0 \\ & \Leftrightarrow {{\left( \dfrac{25}{4} \right)}^{x}}-2.{{\left( \dfrac{10}{4} \right)}^{x}}+1=0 \\ & \Leftrightarrow {{\left( \dfrac{5}{2} \right)}^{2x}}-2.{{\left( \dfrac{5}{2} \right)}^{x}}+1=0 \\ & \Leftrightarrow {{\left( \dfrac{5}{2} \right)}^{x}}=1\Leftrightarrow x=0 \\ \end{aligned} \)

Chọn D.

2.52

\(\begin{aligned} & {{2}^{{{x}^{2}}-6x-\frac{5}{2}}}=16.\sqrt{2} \\ & \Leftrightarrow {{2}^{{{x}^{2}}-6x-\frac{5}{2}}}={{2}^{\frac{9}{2}}} \\ & \Leftrightarrow {{x}^{2}}-6x-\dfrac{5}{2}=\dfrac{9}{2} \\ & \Leftrightarrow {{x}^{2}}-6x-7=0 \\ & \Leftrightarrow \left[ \begin{aligned} & x=-1 \\ & x=7 \\ \end{aligned} \right. \\ \end{aligned} \)

Chọn B