37.
\(a^x.b^{x^2-2}=1\Leftrightarrow b^{x^2-2}=a^{-x}\)
\(\Leftrightarrow x^2-2=-x.log_ba\)
\(\Leftrightarrow x^2+log_ba.x-2=0\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-log_ba\\x_1x_2=-2\end{matrix}\right.\)
Đặt \(log_ba=n>0\)
\(T=\dfrac{4}{n^2}+5n=\dfrac{4}{n^2}+\dfrac{5n}{2}+\dfrac{5n}{2}\ge3\sqrt[3]{\dfrac{100n^2}{4n^2}}=3\sqrt[3]{25}\)
38.
Đặt \(3^x=t\Rightarrow x=log_3t\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}t_1+t_2=m+6\\t_1t_2=243\end{matrix}\right.\)
\(\dfrac{1}{x_1}+\dfrac{1}{x_2}+\dfrac{1}{x_1x_2}=1\)
\(\Rightarrow x_1+x_2+1=x_1x_2\)
\(\Rightarrow log_3t_1+log_3t_2+1=log_3t_1.log_3t_2\)
\(\Rightarrow log_3\left(t_1.t_2\right)+1=log_3t_1.log_3t_2\)
\(\Rightarrow6=log_3t_1.log_3t_2=log_3\left(\dfrac{243}{t_1}\right).log_3t_1\)
\(\Rightarrow log_3t_1\left[5-log_3t_1\right]=6\)
\(\Rightarrow log_3^2t_1-5log_3t_1+6=0\)
\(\Rightarrow\left[{}\begin{matrix}log_3t_1=2\\log_3t_1=3\\\end{matrix}\right.\) \(\Rightarrow\left(t_1;t_2\right)=\left(9;27\right)\)
\(\Rightarrow m+6=9+27\Rightarrow m=30\)
39.
\(9^x+9^{-x}=4+2\left[cos^2\left(nx\right)-sin^2\left(nx\right)\right]\)
\(\Leftrightarrow\left(3^x-3^{-x}\right)^2+2=4+2\left[cos^2\left(nx\right)-sin^2\left(nx\right)\right]\)
\(\Leftrightarrow\left[2cos\left(nx\right)\right]^2=2+2cos^2\left(nx\right)-2sin^2\left(nx\right)\)
\(\Leftrightarrow2cos^2\left(nx\right)+2sin^2\left(nx\right)=2\) (luôn đúng)
Vậy 2 pt đã cho tương đương nhau, nên chúng có cùng tập hợp nghiệm
\(\Rightarrow\) Phương trình có 2023 nghiệm
