Đáp án C.

\(a,log_a1=c\Leftrightarrow a^c=1\Leftrightarrow c=0\Rightarrow log_a1=0\\ b,log_aa=c\Leftrightarrow a^c=a\Leftrightarrow c=1\Rightarrow log_aa=1\\ c,log_aa^c=b\Leftrightarrow a^b=a^c\Leftrightarrow b=c\Rightarrow log_aa^c=c\\ d,a^{log_ab}=c\Leftrightarrow log_ab=log_ac\Leftrightarrow b=c\Rightarrow a^{log_ab}=b\)

a) \(\log_a\left(a^2b\right)=\log_aa^2+\log_ab=2.\log_aa+\log_ab=2.1+2=4\)

b) \(\log_a\dfrac{a\sqrt{a}}{b\sqrt[3]{a}}=\log_a\left(a\sqrt{a}\right)-\log_a\left(b\sqrt[3]{b}\right)=\log_aa^{\dfrac{3}{2}}-\log_ab^{\dfrac{4}{3}}=\dfrac{3}{2}.\log_aa-\dfrac{4}{3}\log_ab=\dfrac{3}{2}.1-\dfrac{4}{3}.2=-\dfrac{7}{6}\)

c) \(\log_a\left(2b\right)+\log_a\left(\dfrac{b^2}{2}\right)=\log_a2+\log_ab+\log_ab^2-\log_a2=\log_ab+2\log_ab=3\log_ab=3.2=6\)

a: \(=log_aa^2+log_ab=2+2=4\)

b: \(log_a\left(\dfrac{a\sqrt{a}}{b\sqrt[3]{b}}\right)=log_aa^{\dfrac{3}{2}}-log_ab^{\dfrac{4}{3}}\)

=3/2-4/3*2

=3/2-8/3

=9/6-16/6=-7/6

c: \(log_a\left(2b\right)+log_a\left(\dfrac{b^2}{2}\right)\)

\(=log_a\left(2b\cdot\dfrac{b^2}{2}\right)=log_a\left(b^3\right)=3\cdot2=6\)

\(log_a\left(a^3b^2\right)=log_aa^3+log_ab^2=3+2\cdot log_ab\)

=>B

\(a,a^{log_ab^{\alpha}}=c\Leftrightarrow log_ac=log_ab^{\alpha}\Leftrightarrow c=b^{\alpha}\Rightarrow a^{log_ab^{\alpha}}=b^{\alpha}\\ a^{\alpha log_ab}=c\Leftrightarrow\alpha log_ab=log_ac\Leftrightarrow log_ab^{\alpha}=log_ac\Leftrightarrow b^{\alpha}=c\Rightarrow a^{\alpha log_ab}=b^{\alpha}\\ \Rightarrow a^{log_ab^{\alpha}}=a^{\alpha log_ab}\)

\(b,a^{log_ab^{\alpha}}=a^{\alpha log_ab}\\ \Rightarrow log_ab^{\alpha}=\alpha log_ab\)

a)    \({\log _c}b = {\log _a}b.{\log _c}a \Leftrightarrow {a^{{{\log }_c}b}} = {a^{{{\log }_a}b.{{\log }_c}a}} \Leftrightarrow {c^{{{\log }_c}b}} = {\left( {{c^{{{\log }_c}a}}} \right)^{{{\log }_a}b}} \Leftrightarrow b = {a^{{{\log }_a}b}} \Leftrightarrow b = b\) (luôn đúng)

Vậy \({\log _c}b = {\log _a}b.{\log _c}a\)

b)    Từ \({\log _c}b = {\log _a}b.{\log _c}a \Leftrightarrow {\log _a}b = \frac{{{{\log }_c}b}}{{{{\log }_c}a}}\)

\(\log_ab^2=2\log_ab=2.3=6\) -> C

Chọn C

\({a^{\frac{1}{2}}} = b \Leftrightarrow {\log _a}b = \frac{1}{2} \Leftrightarrow 2{\log _a}b = 1\)

Chọn B.

xA-yA=2

=>\(x_A-\dfrac{-3}{x_A}=2\)

\(\Leftrightarrow x_A^2+3=2x_A\Leftrightarrow x_A^2-2x_A+3=0\left(loại\right)\)

xB-yB=-6

=>\(x_B-\dfrac{-3}{x_B}=-6\)

\(\Leftrightarrow x_B^2+3=-6x_B\)

=>\(x_B^2+6x_B+3=0\)

=>\(\left[{}\begin{matrix}x=-3+\sqrt{6}\\x=-3-\sqrt{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}y=3+\sqrt{6}\\y=3-\sqrt{6}\end{matrix}\right.\)