Câu 34:
\(2^x\cdot5^{x^2-2x}=1\)
=>\(log_2\left(2^x\cdot5^{x^2-2x}\right)=log_21\)
=>\(x-log_25^{x^2-2x}=0\)
=>\(x-\left(x^2-2x\right)\cdot log_25=0\)
=>\(x\left[1-\left(x-2\right)\cdot log_25\right]=0\)
=>\(\left[{}\begin{matrix}x=0\\1-\left(x-2\right)\cdot log_25=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\\left(x-2\right)\cdot log_25=1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x-2=\dfrac{1}{log_25}=log_52\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=2+log_52\end{matrix}\right.\)
=>Chọn C
Câu 32:
\(\left(3^x-2\right)\left(9^x-3^{x+2}+8\right)=0\)
=>\(\left(3^x-2\right)\left[\left(3^x\right)^2-3^x\cdot9+8\right]=0\)
=>\(\left(3^x-2\right)\left(3^x-1\right)\left(3^x-8\right)=0\)
=>\(\left[{}\begin{matrix}3^x=2\\3^x=1\\3^x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=log_32\\x=log_31\\x=log_38\end{matrix}\right.\)
=>Nghiệm lớn nhất là \(log_38=3\cdot log_32\)
\(P=3log_32-log_32=2log_32\)
=>Chọn C
Câu 33:
Đặt \(t=\left(\sqrt{2}-1\right)^x\left(t>0\right)\)
=>\(\dfrac{1}{t}=\left(\dfrac{1}{\sqrt{2}-1}\right)^x=\left(\sqrt{2}+1\right)^x\)
Phương trình ban đầu sẽ trở thành:
\(t+\dfrac{1}{t}-2\sqrt{2}=0\)
=>\(\dfrac{t^2+1-2\sqrt{2}\cdot t}{t}=0\)
=>\(t^2-2\sqrt{2}t+1=0\)
=>\(\left(t-\sqrt{2}-1\right)\left(t-\sqrt{2}+1\right)=0\)
=>\(\left[{}\begin{matrix}t=\left(\sqrt{2}-1\right)\\t=\sqrt{2}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
=>Chọn C
