\(F=\log_{3-2\sqrt{2}}\left(27^{\log_92}+2^{\log_827}\right)=\log_{3-2\sqrt{2}}\left[\left(3^3\right)^{^{\log_92^2}}+2^{\log_{2^3}3^3}\right]\)

   \(=\log_{3-2\sqrt{2}}\left(3^{\frac{3}{2}\log_32}+2^{\log_23}\right)\)

   \(=\log_{3-2\sqrt{2}}\left(3^{\log_32^{\frac{3}{2}}}+2^{\log_23}\right)\)

   \(=\log_{3-2\sqrt{2}}\left(2^{\frac{3}{2}}+3\right)=\log_{\left(3-2\sqrt{2}\right)^{-1}}\left(3-2\sqrt{2}\right)=-1\)

1.

\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)

\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)

\(=\dfrac{27}{4}log_2a\)

2.

\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)

\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)

\(log_216=log_22^4=4\)

\(log_32187=log_33^7=7\)

\(log_{10}\dfrac{1}{100}=log_{10}10^{-2}=-2\)

\(log10000=log10^4=4\)

\(9^{log_312}=3^{2log_312}=3^{log_3144}=144\)

\(8^{log_25}=2^{3log_25}=2^{log_2125}=125\)

\(\left(\dfrac{1}{25}\right)^{log_5\dfrac{1}{3}}=5^{-2log_5\dfrac{1}{3}}=5^{log_59}=9\)

\(\left(\dfrac{1}{4}\right)^{log_23}=2^{-2log_23}=2^{log_2\dfrac{1}{9}}=\dfrac{1}{9}\)

a) \(A=\log_{5^{-2}}5^{\frac{5}{4}}=-\frac{1}{2}.\frac{5}{4}.\log_55=-\frac{5}{8}\)

b) \(B=9^{\frac{1}{2}\log_22-2\log_{27}3}=3^{\log_32-\frac{3}{4}\log_33}=\frac{2}{3^{\frac{3}{4}}}=\frac{2}{3\sqrt[3]{3}}\)

c) \(C=\log_3\log_29=\log_3\log_22^3=\log_33=1\)

d) Ta có \(D=\log_{\frac{1}{3}}6^2-\log_{\frac{1}{3}}400^{\frac{1}{2}}+\log_{\frac{1}{3}}\left(\sqrt[3]{45}\right)\)

                   \(=\log_{\frac{1}{3}}36-\log_{\frac{1}{3}}20+\log_{\frac{1}{3}}45\)

                   \(=\log_{\frac{1}{3}}\frac{36.45}{20}=\log_{3^{-1}}81=-\log_33^4=-4\)

\(a,log_272-\dfrac{1}{2}\left(log_23+log_227\right)\\ =log_272-\dfrac{1}{2}log_2\left(3\cdot27\right)\\ =log_272-log_2\left(81\right)^{\dfrac{1}{2}}\\ =log_272-log_29\\ =log_2\dfrac{72}{9}\\ =log_28\\ =3\)

\(b,5^{log_240-log_25}\\ =5^{log_2\dfrac{40}{5}}\\ =5^{log_28}\\ =5^3\\ =125\)

\(c,3^{2+log_92}\\ =3^{log_9\left(81\cdot2\right)}\\ =3^{\dfrac{1}{2}log_3162}\\ =\left(162\right)^{\dfrac{1}{2}}\\ =\sqrt{162}\\ =9\sqrt{2}\)

Chọn 2 làm cơ số, ta có :

\(A=\log_616=\frac{\log_216}{\log_26}=\frac{4}{1=\log_23}\)

Mặt khác :

\(x=\log_{12}27=\frac{\log_227}{\log_212}=\frac{3\log_23}{2+\log_23}\)

Do đó : \(\log_23=\frac{2x}{3-x}\) suy ra \(A=\frac{4\left(3-x\right)}{3+x}\)

b) Ta có :

\(B=\frac{lg30}{lg125}=\frac{lg10+lg3}{3lg\frac{10}{2}}=\frac{1+lg3}{3\left(1-lg2\right)}=\frac{1+a}{3\left(1-b\right)}\)

c) Ta có :

\(C=\log_65+\log_67=\frac{1}{\frac{1}{\log_25}+\frac{1}{\log_35}}+\frac{1}{\frac{1}{\log_27}+\frac{1}{\log_37}}\)

Ta tính \(\log_25,\log_35,\log_27,\log_37\) theo a, b, c .

Từ : \(a=\log_{27}5=\log_{3^3}5=\frac{1}{3}\log_35\)

Suy ra \(\log_35=3a\) do đó :

                                     \(\log_25=\log_23.\log35=3ac\)

Mặt khác : \(b=\log_87=\log_{2^3}7=\frac{1}{3}\log_27\) nên \(\log_27=3b\)

Do đó : \(\log_37=\frac{\log_27}{\log_23}=\frac{3b}{c}\)

Vậy : \(C=\frac{1}{\frac{1}{3ac}+\frac{1}{3a}}+\frac{1}{\frac{1}{3b}+\frac{c}{3b}}=\frac{3\left(ac+b\right)}{1+c}\)

d) Điều kiện : \(a>0;a\ne0;b>0\)

Từ giả thiết \(\log_ab=\sqrt{3}\) suy ra \(b=a^{\sqrt{3}}\). Do đó :

\(\frac{\sqrt{b}}{a}=a^{\frac{\sqrt{3}}{2}-1};\frac{\sqrt[3]{b}}{\sqrt{a}}=a^{\frac{\sqrt{3}}{3}-\frac{1}{2}}=a^{\frac{\sqrt{3}}{3}\left(\frac{\sqrt{3}}{2}-1\right)}\)

Từ đó ta tính được :

\(A=\log_{a^{\alpha}}a^{\frac{-\sqrt{3}}{3}\alpha}=\log_{a^{\alpha}}\left(a^{\alpha}\right)^{\frac{-\sqrt{3}}{3}}=\frac{-\sqrt{3}}{3}\) với \(\alpha=\frac{\sqrt{3}}{2}-1\)

Ta có :

\(\sqrt{2}=2^{\frac{1}{2}}\)

\(\left(2^3\right)^{\log_{64}\frac{5}{4}}=2^{3\log_{2^6}\frac{5}{4}}=2^{\frac{1}{2}\log_2\frac{5}{4}}=2^{\log_2\sqrt{\frac{5}{4}}}=\sqrt{\frac{5}{4}}=\left(\frac{5}{4}\right)^{\frac{1}{2}}\)

\(2^{3^{\log_92}}=2^{3^{\frac{1}{2}\log_32}}=2^{3^{\log_3\sqrt{2}}}=2^{\sqrt{2}}\)

Mà : \(\sqrt{2}>\frac{\pi}{6}>\frac{1}{2}\Rightarrow2^{\sqrt{2}}>2^{\frac{\pi}{6}}>2^{\frac{1}{2}}\)

                            \(\Leftrightarrow2^{3^{\log_92}}>2^{\frac{\pi}{6}}>\sqrt{2}\)  (1)

Mặt khác : \(2>\frac{5}{4}\Rightarrow2^{\frac{1}{2}}>\left(\frac{5}{4}\right)^{\frac{1}{2}}\) hay \(\sqrt{2}>\left(2^3\right)^{\log_{64}\frac{5}{4}}\)  (2)

Từ (1) và (2) : \(2^{3^{\log_92}}>2^{\frac{\pi}{6}}>\sqrt{2}>\left(2^3\right)^{\log_{64}\frac{5}{4}}\)

Vậy thứ tự giảm dần là :

\(2^{3^{\log_92}};2^{\frac{\pi}{6}};\sqrt{2};\left(2^3\right)^{\log_{64}\frac{5}{4}}\)

a) Áp dụng bất đẳng thức Cauchy cho các số dương, ta có :

\(\log_23+\log_32>2\sqrt{\log_23.\log_32}=2\sqrt{1}=2\)

Không xảy ra dấu "=" vì \(\log_23\ne\log_32\)

Mặt khác, ta lại có :

\(\log_23+\log_32<\frac{5}{2}\Leftrightarrow\log_23+\frac{1}{\log_23}-\frac{5}{2}<0\)

                             \(\Leftrightarrow2\log^2_23-5\log_23+2<0\)

                            \(\Leftrightarrow\left(\log_23-1\right)\left(\log_23-2\right)<0\) (*)

Hơn nữa, \(2\log_23>2\log_22>1\) nên \(2\log_23-1>0\)

Mà \(\log_23<\log_24=2\Rightarrow\log_23-2<0\)

Từ đó suy ra (*) luôn đúng. Vậy \(2<\log_23+\log_32<\frac{5}{2}\)

b) Vì \(a,b\ge1\) nên \(\ln a,\ln b,\ln\frac{a+b}{2}\) không âm. 

Áp dụng bất đẳng thức Cauchy ta có

\(\ln a+\ln b\ge2\sqrt{\ln a.\ln b}\)

Suy ra 

\(2\left(\ln a+\ln b\right)\ge\ln a+\ln b+2\sqrt{\ln a\ln b}=\left(\sqrt{\ln a}+\sqrt{\ln b}\right)^2\)

Mặt khác :

\(\frac{a+b}{2}\ge\sqrt{ab}\Rightarrow\ln\frac{a+b}{2}\ge\frac{1}{2}\left(\ln a+\ln b\right)\)

Từ đó ta thu được :

\(\ln\frac{a+b}{2}\ge\frac{1}{4}\left(\sqrt{\ln a}+\sqrt{\ln b}\right)^2\)

hay \(\frac{\sqrt{\ln a}+\sqrt{\ln b}}{2}\le\sqrt{\ln\frac{a+b}{2}}\)

c) Ta chứng minh bài toán tổng quát :

\(\log_n\left(n+1\right)>\log_{n+1}\left(n+2\right)\) với mọi n >1

Thật vậy, 

\(\left(n+1\right)^2=n\left(n+2\right)+1>n\left(n+2\right)>1\) 

suy ra :

\(\log_{\left(n+1\right)^2}n\left(n+2\right)<1\Leftrightarrow\frac{1}{2}\log_{n+1}n\left(n+2\right)<1\)

                                  \(\Leftrightarrow\log_{n+1}n+\log_{\left(n+1\right)}n\left(n+2\right)<2\)

Áp dụng bất đẳng thức Cauchy ta có :

\(2>\log_{\left(n+1\right)}n+\log_{\left(n+1\right)}n\left(n+2\right)>2\sqrt{\log_{\left(n+1\right)}n.\log_{\left(n+1\right)}n\left(n+2\right)}\)

Do đó ta có :

\(1>\log_{\left(n+1\right)}n.\log_{\left(n+1\right)}n\left(n+2\right)\) và \(\log_n\left(n+1>\right)\log_{\left(n+1\right)}\left(n+2\right)\) với mọi n>1