Có: \(f\left(3\right)=5\cdot3+1=15+1=16\)

       \(f\left(7\right)=7-3\cdot7=7-21=-14\)

=> \(f\left(3\right)\cdot g\left(7\right)=16\cdot\left(-14\right)=-224\)

\(x^2+5y^2-2xy+4y+1=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(4y^2+4y+1\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(2y+1\right)^2=0\)

\(\Leftrightarrow\begin{cases}x-y=0\\2y+1=0\end{cases}\)\(\Leftrightarrow x=y=-\frac{1}{2}\)

\(\frac{x-5}{x-6}=\frac{x-1}{x+2}\)

=> \(\left(x-5\right)\left(x+2\right)=\left(x-6\right)\left(x-1\right)\)

=> \(x^2+2x-5x-10=x^2-x-6x+6\)

=> \(x^2+2x-5x-x^2+x+6x=6+10\)

\(=>4x=16\)

=> \(x=4\)

\(9\left(x-y\right)^2-4\left(x+y\right)^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\)

\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)

\(=\left(x-5y\right)\left(5x-y\right)\)

\(x-2\sqrt{x}=0\left(ĐK:x\ge0\right)\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)

Vậy x={0;4}

Có: \(x+y=2\)

=> \(x^2+2xy+y^2=4\)

=>\(2xy=4-10\)

=> \(2xy=-6\)

=>\(xy=-3\)

\(x^3+y^3=\left(x+y\right)\left(x^2-xy+y^2\right)=2\cdot\left(10+3\right)=36\)

\(\left(\frac{1}{2a-b}+\frac{3b}{b^2-4a^2}-\frac{2}{2a+b}\right):\left(1+\frac{4a^2+b^2}{4a^2-b^2}\right)\left(ĐK:2a\ne\pm b\right)\)

\(=\left(\frac{1}{2a-b}-\frac{3b}{\left(2b-b\right)\left(2a+b\right)}-\frac{2}{2a+b}\right):\frac{4a^2-b^2+4a^2+b^2}{\left(2a-b\right)\left(2a+b\right)}\)

\(=\frac{2a+b-3b-2\left(2a-b\right)}{\left(2a-b\right)\left(2a+b\right)}\cdot\frac{\left(2a-b\right)\left(2a+b\right)}{8a^2}\)

\(=\frac{2a+b-3b-4a+2b}{8a^2}=\frac{-2a}{8a^2}=-\frac{1}{4a}\)

Số số hạng của A là:2015-1+1=2015(số)

Tổng dãy A là:

(2015+1)x2015:2=2031120