Phương trình (1+căn bậc hai của 2)^x +(1-2a)(căn bậc hai của 2 -1)^x

* Hướng dẫn giải

Chọn B

$$\begin{gathered} \mathrm{Đặt} {\left(1+\sqrt{2}\right)}^{\mathrm{x}}=\mathrm{t}\Rightarrow {\left(\sqrt{2}-1\right)}^{\mathrm{x}}=\frac{1}{\mathrm{t}} \\ \Rightarrow \mathrm{t}+(1-2\mathrm{a})\frac{1}{\mathrm{t}}-4=0\iff {\mathrm{t}}^2-4\mathrm{t}+1-2\mathrm{a}=0 \left(1\right) \\ \mathrm{PT}(*) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{phân} \mathrm{biệt} \mathrm{khi} \mathrm{PT}\left(1\right) \mathrm{có} 2 \mathrm{nghiệm} \mathrm{phân} \mathrm{biệt} \mathrm{dương} \\ \iff \left\{\begin{array}{l}∆\text{'}>0\\ \mathrm{S}>0\\ \mathrm{P}>0\end{array}\iff \left\{\begin{array}{l}4+2\mathrm{a}-1>0\\ 4>0 (\mathrm{luôn} \mathrm{đúng})\\ 1-2\mathrm{a}>0\end{array}\iff \left\{\begin{array}{l}\mathrm{a}>\frac{-3}{2}\\ \mathrm{a}<\frac{1}{2}\end{array}\right.\right.\right. \\ \iff \frac{-3}{2}<\mathrm{a}<\frac{1}{2} \\ \mathrm{khi} \mathrm{đó}\left\{\begin{array}{l}{\left(1+\sqrt{2}\right)}^{{\mathrm{x}}_1}={\mathrm{t}}_1\\ {\left(1+\sqrt{2}\right)}^{{\mathrm{x}}_2}={\mathrm{t}}_2\end{array}\right.\iff \left\{\begin{array}{l}{\mathrm{x}}_1={\log }_{1+\sqrt{2}} {\mathrm{t}}_1\\ {\mathrm{x}}_2={\log }_{1+\sqrt{2}} {\mathrm{t}}_2\end{array}\right. \\ \mathrm{Theo} \mathrm{ĐB}: {\mathrm{x}}_1-{\mathrm{x}}_2={\log }_{1+\sqrt{2}}3 \\ \iff {\log }_{1+\sqrt{2}} {\mathrm{t}}_1-{\log }_{1+\sqrt{2}} {\mathrm{t}}_2={\log }_{1+\sqrt{2}}3 \\ \iff \frac{{\mathrm{t}}_1}{{\mathrm{t}}_2}=3\iff {\mathrm{t}}_1=3{\mathrm{t}}_2 \\ \mathrm{Theo} \mathrm{Vi}-\mathrm{ét}: \left\{\begin{array}{l}{\mathrm{t}}_1+{\mathrm{t}}_2=4 \\ {\mathrm{t}}_1{\mathrm{t}}_2=1-2\mathrm{a}\end{array}\right.. \mathrm{Thay} {\mathrm{t}}_1=3{\mathrm{t}}_2 \mathrm{ta} \mathrm{tìm} \mathrm{được}\left\{\begin{array}{l}{\mathrm{t}}_1=3\\ {\mathrm{t}}_2=1\end{array}\right. \\ \Rightarrow 1-2\mathrm{a}=3\iff \mathrm{a}=-1\in \left(\frac{-3}{2};0\right)\left(TMĐK\right) \end{gathered}$$