\(a+b=12\)

\(\Rightarrow a=12-b\)

và \(b=12-a\)

\(a+b=12\Rightarrow\orbr{\begin{cases}a=12-b\\b=12-a\end{cases}}\)

Theo AM - GM :

\(a+b\ge2\sqrt{ab}\)

\(\Leftrightarrow ab\le\dfrac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow a+b+2ab\le\dfrac{a^2+b^2+2ab+2a+2b}{2}\)

\(\Leftrightarrow a^2+b^2+2ab+2a+2b\ge24\)

\(\Leftrightarrow\left(a+b+6\right)\left(a+b-4\right)\ge0\)

\(\Leftrightarrow A\ge4\) (do a,b dương)

Dấu"=" \(\Leftrightarrow a=b=2\)

Ta có: \(A+B+C=0\)

\(\Leftrightarrow3x^2y+5xy^2-2xy+1+2x^2y-7xy^2+6xy-8-5x^2y+4xy^2-4xy+12=0\)

\(\Leftrightarrow2xy^2+5=0\)

\(\Leftrightarrow2x\cdot\left(-2\right)^2+5=0\)

\(\Leftrightarrow8x+5=0\)

\(\Leftrightarrow8x=-5\)

hay \(x=-\dfrac{5}{8}\)

Vậy: \(x=-\dfrac{5}{8}\)

\(A^2+B^2=\left(A+B\right)^2-2AB=5\)

\(A^3+B^3=\left(A+B\right)^3-3AB\left(A+B\right)=9\)

\(A^5+B^5=\left(A^2+B^2\right)\left(A^3+B^3\right)-\left(AB\right)^2\left(A+B\right)=5.9-2^2.3=...\)

B.

\(A^2+B^2=\left(A+B\right)^2-2AB=2\)

\(A^6+B^6=\left(A^2\right)^3+\left(B^2\right)^3=\left(A^2+B^2\right)^3-3\left(AB\right)^2\left(A^2+B^2\right)=2^3-3.1^2.2=...\)

Ta có: \(A^2+B^2=\left(A+B\right)^2-2AB=3^2-2.2=5\)

\(A^5+B^5=\left(A^3+B^3\right)\left(A^2+B^2\right)-A^2B^2\left(A+B\right)=\left(A+B\right)\left(A^2-AB+B^2\right)\left(A^2+B^2\right)-A^2B^2\left(A+B\right)=3\left(5-2\right).5-2^2.3=33\)

\(log_65=\dfrac{1}{log_56}=\dfrac{1}{log_52+log_53}=\dfrac{1}{a+b}\)

=>Chọn B

Ta có: a+b=5

\(\Leftrightarrow\left(a+b\right)^2=25\)

\(\Leftrightarrow a^2+b^2+2ab=25\)

\(\Leftrightarrow2ab=16\)

hay ab=8

Ta có: \(a^3+b^3\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(=5^3-3\cdot8\cdot5=5\)