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\(P=\dfrac{1}{log_a\dfrac{a}{b}}+log_bb-log_ba=\dfrac{1}{1-log_ab}+1-log_ba\)
\(=\dfrac{log_ba}{log_ba-1}+1-log_ba\)
Đặt \(log_ba=x\Rightarrow x\ge2\)
\(P=f\left(x\right)=\dfrac{x}{x-1}+1-x\)
\(f'\left(x\right)=\dfrac{-1}{\left(x-1\right)^2}-1< 0\) \(\Rightarrow\) hàm nghịch biến
\(\Rightarrow P\) chỉ tồn tại max (tại \(x=2\)), ko tồn tại min
Đề sai
\(P=\frac{1}{2}log_{\frac{a}{b}}a-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2log_a\frac{a}{b}}-4log_a\left(a+\frac{b}{4}\right)=\frac{1}{2\left(1-log_ab\right)}-4log_a\left(a+\frac{b}{4}\right)\)
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 \(\left\{{}\begin{matrix}x=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}log_ab=\frac{1}{2}\\a=\frac{b}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=b^2\\a=\frac{b}{4}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{16}\\b=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow S=\frac{5}{16}\)
\(P=log_{\dfrac{\sqrt{a}}{b}}a+log_{\dfrac{\sqrt{a}}{b}}\sqrt[3]{b}=log_{\dfrac{\sqrt{a}}{b}}a+\dfrac{1}{3}log_{\dfrac{\sqrt{a}}{b}}b\)
\(=\dfrac{1}{log_a\dfrac{\sqrt{a}}{b}}+\dfrac{1}{3.log_b\dfrac{\sqrt{a}}{b}}=\dfrac{1}{log_a\sqrt{a}-log_ab}+\dfrac{1}{3\left(log_b\sqrt{a}-log_bb\right)}\)
\(=\dfrac{1}{\dfrac{1}{2}-2}+\dfrac{1}{3\left(\dfrac{1}{4}-1\right)}=-\dfrac{10}{9}\)
\(log_{a^2}\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}log_a\left(\dfrac{a^3}{\sqrt[5]{b^3}}\right)=\dfrac{1}{2}\left[log_aa^3-log_a\sqrt[5]{b^3}\right]=\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)\)
\(\Rightarrow\dfrac{1}{2}\left(3-\dfrac{3}{5}log_ab\right)=3\)
\(\Rightarrow log_ab=-5\)
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=logab
Do b<a<1 => t=logab >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\(\left(\dfrac{3}{2}\right)\) = \(\dfrac{9}{2}\)