36.
Đặt \(\left(4+\sqrt{15}\right)^x=t>0\Rightarrow\left(4-\sqrt{15}\right)^x=\dfrac{1}{t}\)
Ta trở thành:
\(t+\dfrac{1}{t}=62\Rightarrow t^2-62t+1=0\)
\(\Rightarrow t_1+t_2=1\) theo định lý Viet
\(\Rightarrow\left(4+\sqrt{15}\right)^{x_1}.\left(4+\sqrt{15}\right)^{x_2}=t_1t_2=1\)
\(\Rightarrow\left(4+\sqrt{15}\right)^{x_1+x_2}=1\)
\(\Rightarrow x_1+x_2=log_{4+\sqrt{15}}1=0\)
37.
Đặt \(\left(\sqrt{6+\sqrt{35}}\right)^x=t>0\Rightarrow\left(\sqrt{6-\sqrt{35}}\right)^x=\dfrac{1}{t}\)
\(\Rightarrow t+\dfrac{1}{t}=12\Rightarrow t^2-12t+1=0\)
\(\Rightarrow t=6\pm\sqrt[]{35}\)
\(\Rightarrow\left[{}\begin{matrix}\left(\sqrt{6+\sqrt{35}}\right)^x=6+\sqrt{35}=\left(\sqrt{6+\sqrt{35}}\right)^2\\\left(\sqrt{6+\sqrt{35}}\right)^x=6-\sqrt{35}=\left(\sqrt{6+\sqrt{35}}\right)^{-2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) \(\Rightarrow\) tích các nghiệm là -4
38.
Đặt \(\left(4+\sqrt{15}\right)^x=t>0\Rightarrow\left(4-\sqrt{15}\right)^x=\dfrac{1}{t}\)
Pt trở thành:
\(t+\dfrac{1}{t}=62\Leftrightarrow t^2-62t+1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=31+8\sqrt{15}=\left(4+\sqrt{15}\right)^2\\t=31-8\sqrt{15}=\left(4+\sqrt{15}\right)^{-2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(4+\sqrt{15}\right)^x=\left(4+\sqrt{15}\right)^2\\\left(4+\sqrt{15}\right)^x=\left(4+\sqrt{15}\right)^{-2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\) \(\Rightarrow\) tập nghiệm \(\left\{-2;2\right\}\)
39.
Đặt \(\left(2+\sqrt{3}\right)^x=t>0\Rightarrow\left(2-\sqrt{3}\right)^x=\dfrac{1}{t}\)
\(\Rightarrow t+\dfrac{1}{t}=4\Rightarrow t^2-4t+1=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2+\sqrt{3}\\t=2-\sqrt{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left(2+\sqrt{3}\right)^x=2+\sqrt{3}\\\left(2+\sqrt{3}\right)^x=2-\sqrt{3}=\left(2+\sqrt{3}\right)^{-1}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\) \(\Rightarrow\) tổng các nghiệm \(1-1=0\)
Em cần câu nào trong 4 câu này nhỉ?
