Câu 46: ĐKXĐ: x>0
\(\log_2^2\left(2x\right)-3\cdot\log_2x-2=0\)
=>\(\left\lbrack\log_22x\right\rbrack^2-3\cdot\log_2x-2=0\)
=>\(\left\lbrack1+\log_2x\right\rbrack^2-3\cdot\log_2x-2=0\)
=>\(\left(\log_2x\right)^2+2\cdot\log_2x+1-3\cdot\log_2x-2=0\)
=>\(\left(\log_2x\right)^2-\log_2x-1=0\)
Đặt \(a=\log_2x\)
Phương trình sẽ trở thành: \(a^2-a-1=0\)
=>\(a^2-a+\frac14=\frac54\)
=>\(\left(a-\frac12\right)^2=\frac54\)
=>\(\left[\begin{array}{l}a-\frac12=\frac{\sqrt5}{2}\\ a-\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\frac{\sqrt5+1}{2}\\ a=\frac{-\sqrt5+1}{2}\end{array}\right.\)
=>\(\left[\begin{array}{l}x=2^{\frac{1+\sqrt5}{2}}\\ x=2^{\frac{1-\sqrt5}{2}}\end{array}\right.\)
mà \(x=2^{\frac{a+\sqrt{b}}{c}}\)
nên a=1; b=5; c=2
\(a+b+c^2=1+5+2^2=10\)
=>Chọn A
