\(P=\frac{1}{2}log_{\frac{a}{b}}a-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2log_a\frac{a}{b}}-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2\left(1-log_ab\right)}-4log_a\left(a+\frac{b}{4}\right)\)

Ta có: \(a+\frac{b}{4}\ge2\sqrt{\frac{ab}{4}}=\sqrt{ab}\)

\(\Rightarrow log_a\left(a+\frac{b}{4}\right)\le log_a\sqrt{ab}\) (do \(0< a< 1\))

\(\Rightarrow P\ge\frac{1}{2\left(1-log_ab\right)}-4log_a\sqrt{ab}=\frac{1}{2\left(1-log_ab\right)}-2\left(1+log_ab\right)\)

Đặt \(log_ab=x\Rightarrow0< x< 1\) \(\Rightarrow P\ge\frac{1}{2\left(1-x\right)}-2\left(1+x\right)\)

Xét hàm \(f\left(x\right)=\frac{1}{2\left(1-x\right)}-2\left(1+x\right)\) với \(0< x< 1\)

\(f'\left(x\right)=\frac{1}{2\left(1-x\right)^2}-2=0\Leftrightarrow\frac{1-4\left(1-x\right)^2}{2\left(1-x\right)^2}=0\Rightarrow x=\frac{1}{2}\)

Từ BBT ta thấy \(f\left(x\right)_{min}=f\left(\frac{1}{2}\right)=-2\)

\(\Rightarrow P\ge-2\Rightarrow P_{min}=-2\) khi \(\left\{{}\begin{matrix}x=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}log_ab=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\a=\frac{b}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{16}\\b=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow S=\frac{5}{16}\)

\(A=\frac{1}{x-2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)

a)\(A=\frac{1}{x-2}+\frac{1}{x-2}+\frac{x^2+1}{x^2-4}\)

\(=\frac{x-2+x-2}{x^2-4}+\frac{x^2+1}{x^2-4}\)

\(=\frac{x^2+2x-3}{x^2-4}\)

đầu bài sai rồi bạn ơi bạn cho x=0 thì \(A=\frac{3}{4}\)là số dương rồi 

Ta có: 

\(\left(b-\dfrac{1}{2}\right)^2\ge0\) <=> \(b^2-b+\dfrac{1}{4}\ge0\) <=>\(b-\dfrac{1}{4}\le b^2\)

Mà : 

a<1 => \(log_a\left(b-\dfrac{1}{4}\right)\ge log_ab^2=2log_ab\)

P=\(log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}log_{\dfrac{a}{b}}b=log_a\left(b-\dfrac{1}{4}\right)-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\ge2log_ab-\dfrac{1}{2}.\dfrac{log_ab}{1-log_ab}\)

Đặt t=log_ab

Do b<a<1 => t=log_ab >1

Khi đó \(P\ge2t+\dfrac{t}{2t-2}=f\left(t\right)\). Khảo sát f(t) trên (1;+\(\infty\)) ta đc

P\(\ge\)f(t) \(\ge\) f\(\left(\dfrac{3}{2}\right)\) = \(\dfrac{9}{2}\)

Chọn đáp án C.