Question

log_x2 + log₂x ≥ \(5/2\)

Answer 1

ĐKXĐ: x>0

Đặt \(a=log_x2\left(a>=0\right)\)

BPT sẽ trở thành \(a+\dfrac{1}{a}>=\dfrac{5}{2}\)

=>\(\dfrac{a^2+1}{a}>=\dfrac{5}{2}\)

=>\(2\left(a^2+1\right)>=5a\)

=>\(2a^2-5a+2>=0\)

=>\(\left(2a-1\right)\left(a-2\right)>=0\)

=>\(\left[{}\begin{matrix}a>=2\\a< =\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}a>=2\\0< a< =\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}log_x2>=2\\0< log_x2< =\dfrac{1}{2}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>=\sqrt{2}\\0< x< =4\end{matrix}\right.\)