Question

Xét các số thực \(x\), \(y\) thỏa mãn \(\log_2\left(x+\sqrt{x^2+1}\right)+\log_2\left(y+\sqrt{y^2+1}\right)=4\). Kí hiệu \(m\) là giá trị nhỏ nhất \(P=x+y\). Mệnh đề nào sau đây đúng?

A. \(m\in\left(4;5\right)\)

B. \(m\in\left(3;\dfrac{7}{2}\right)\)

C. \(m\in\left(\dfrac{7}{2};4\right)\)

D. \(m\in\left(\dfrac{5}{2};3\right)\)

Answer 1

\(log_2\left(x+\sqrt{x^2+1}\right)+log_2\left(y+\sqrt{y^2+1}\right)=4\)

\(\Leftrightarrow log_2\left[\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)\right]=4\)

\(\Leftrightarrow\left(x+\sqrt{x^2+1}\right)\left(y+\sqrt{y^2+1}\right)=16\)

Đặt \(\left\{{}\begin{matrix}x+\sqrt{x^2+1}=a>0\\y+\sqrt{y^2+1}=b>0\end{matrix}\right.\) \(\Rightarrow ab=16\)

\(\sqrt{x^2+1}=a-x\Rightarrow x^2+1=a^2-2ax+x^2\)

\(\Rightarrow2ax=a^2-1\Rightarrow x=\dfrac{a^2-1}{2a}\)

Tương tự \(\Rightarrow y=\dfrac{b^2-1}{2b}\)

\(P=x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a+b}{2}-\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(=\dfrac{a+b}{2}-\dfrac{1}{2}\left(\dfrac{a+b}{ab}\right)=\dfrac{a+b}{2}-\dfrac{a+b}{32}\) (do \(ab=16\))

\(=\dfrac{15}{32}\left(a+b\right)\ge\dfrac{15}{32}.2\sqrt{ab}=\dfrac{15}{32}.2\sqrt{16}=\dfrac{15}{4}\)

\(\Rightarrow m=\dfrac{15}{4}=3,75\) (C đúng)