\(B=\left(\log b_a+\log_ba+2\right)\left(\log b_a-\log b_{ab}\right)-1=\left(\log b_a+\frac{1}{\log b_a}+2\right)\left(\log b_a.\log_ba-\left(\log_{ab}b.\log_ba\right)\right)-1\)

   \(=\frac{\log^2_ab+2\log_ab+1}{\log_ab}\left(1-\log_{ab}a\right)-1=\frac{\left(\log_ab+1\right)^2}{\log_ab}\left(1-\frac{1}{\log_aab}\right)-1\)

  \(=\frac{\left(\log_ab+1\right)^2}{\log_ab}\left(1-\frac{1}{1+\log_ab}\right)-1=\frac{\left(\log_ab+1\right)^2}{\log_ab}.\frac{\log_ab}{1+\log_ab}-1=\log_ab+1-1=\log_ab\)

\(=\left(\log_ab+\log_ba+2\right)\left(1-\log_{ab}a\right)-1\)

\(=\left(\log_ab+\log_ba+2\right)\left(1-\frac{1}{1+\log_ab}\right)-1\)

\(=\frac{1}{1+\log_ab}\left(\log_ab+\log_ba+2\right)-1\)

\(=\frac{1}{1+\log_ab}\left[\left(\log_ab+\log_ba+2\right)-1-\log_ab\right]\)

\(=\frac{1}{1+\log_ab}\left(\log_ab+\log^2_ba\right)=\log_ab\)

\(log_{a^4}b^4.log_ba^5=\dfrac{1}{4}.4.log_ab.5.log_ba=5.log_ab.log_ba=5\)

\(log_{a^3}b^2.log_ba^4=\dfrac{1}{3}.2.log_ab.4.log_ba=\dfrac{8}{3}.log_ab.log_ba=\dfrac{8}{3}\)

\(log_{a^{15}}b^7.log_{b^{49}}a^{30}=\dfrac{1}{15}.7.log_ab.\dfrac{1}{49}.30.log_ba=\dfrac{2}{7}log_ab.log_ba=\dfrac{2}{7}\)

\(log_{a^{2021}}b^{2020}.log_{b^{4040}}a^{6063}=\dfrac{1}{2021}.2020.log_ab.\dfrac{1}{4040}.6063.log_ba=\dfrac{3}{2}\)

Lời giải:

Đặt \(\log_ab=x\Rightarrow \log_ba=\frac{1}{x}\)

a)

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

\(\Leftrightarrow A=(1+\frac{1}{x^2}+2x)(x-\frac{1}{x})=\left(1+\frac{1}{x}\right)^2(x-\frac{1}{x})\)

\(\Leftrightarrow A=(1+\log_ba)^2(\log_ab-\log_ba)\)

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b) Điều kiện: \(x>0\)

Có \(1=\log_{ab}b.\log_b(ab)=\log_{ab}b(\log_ba+\log_bb)=\log_{ab}b(\frac{1}{x}+1)\)

\(\Rightarrow \log_{ab}b=\frac{x}{x+1}\)

Như vậy:

\(B=\sqrt{x+\frac{1}{x}+2}(x-\frac{x}{x+1})\sqrt{x}\)

\(\Leftrightarrow B=\sqrt{x^2+1+2x}(x-\frac{x}{x+1})=|x+1|.\frac{x^2}{x+1}\)

\(=(x+1)\frac{x^2}{x+1}=x^2=\log_a^2b\) (do \(x>0)\)

ta có \(\left(log^b_a+log^a_b+2\right)\left(log^b_a-log_{ab}^b\right).log_b^a-1=\left(log^b_a+log^a_b+2\right)\left(log^b_a.log_b^a-log_{ab}^b.log_b^a\right)-1=\left(log^b_a+log^a_b+2\right)\left(1-\frac{1}{log_b^{ba}}log_b^a\right)-1=\left(log^b_a+log^a_b+2\right)\left(1-\frac{1}{1+log^a_b}log^a_b\right)-1=\left(log^b_a+log^a_b+2\right)\frac{1}{1+log^a_b}-1=\left(log^a_b+\frac{1}{log^a_b}+2\right)\frac{1}{1+log^a_b}-1=\frac{\left(1+log^a_b\right)^2}{log^a_b}\frac{1}{1+log^a}-1=\frac{1+log^a_b}{log_b^a}-1=\frac{1}{log_b^a}\)

 ta có:

\(\left(log^b_a+\frac{1}{log^b_a}+2\right)\left(log^b_a-\frac{1}{log^{ab}_a}\right)log^a_b-1\)\(=\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(log^b_a-\frac{1}{1+log^b_a}\right)log^a_b-1\)\(=\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(1-\frac{log^a_b}{1+log^b_a}\right)-1\)\(==\frac{\left(log^b_a+1\right)^2}{log^b_a}\left(\frac{1}{1+log^b_a}\right)-1=\frac{1+log^b_a}{log^b_a}-1=\frac{1}{log^b_a}\)

\(A=log_2\left(x^3-x\right)-log_2\left(x+1\right)-log_2\left(x-1\right)\)

\(=log_2\left(\dfrac{x^3-x}{x+1}\right)-log_2\left(x-1\right)\)

\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{x+1}\right)-log_2\left(x-1\right)\)

\(=log_2\left(\dfrac{x\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)=log_2x\)

1.\(\dfrac{log_ac}{log_{ab}c}=log_ac.log_c\left(ab\right)=log_ac.\left(log_ca+log_cb\right)=log_ac.log_ca+log_ac.log_cb=\dfrac{log_ac}{log_ac}+\dfrac{log_cb}{log_ca}=1+log_ab\)

2. \(log_{ax}bx=\dfrac{log_abx}{log_aax}=\dfrac{log_ab+log_ax}{log_aa+log_ax}=\dfrac{log_ab+log_ax}{1+log_ax}\)

3. \(\dfrac{1}{log_ax}+\dfrac{1}{log_{a^2}x}+...+\dfrac{1}{log_{a^n}x}=log_xa+log_xa^2+...+log_xa^n\)

\(=log_xa+2log_xa+...+n.log_xa=log_xa+2log_xa+...+n.log_xa\)

\(=log_xa.\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2}log_xa=\dfrac{n\left(n+1\right)}{2.log_ax}\)

 ta có:

\(log^{\left(2a^2\right)}_2+\left(log_2^a\right)a^{log_a^{\left(log^a_1+1\right)}}+\frac{1}{2}log^2_2a^4=log_2^2+log_2^{a^2}+log_2^a\left(log^a_2+1\right)+\frac{1}{2}log^2_2a^4\)

\(=1+2log^a_2+log^a_2\left(1+log^a_2\right)+2log^2a_2\)

\(=3log^2_2a+3log^a_2+1\)

\(R=\log_22x^2+\left(\log_2x\right)x^{\log_x\left(\log_2x+1\right)}+\frac{1}{2}\log^2_4x^4\)

    \(=1+2\log_2x+\left(\log_2x\right)\left(\log_2x+1\right)+2\log^2_2x\)

    \(=3\log^2_2x+3\log_2x+1\)