\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{49.50}\)

\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+........+\frac{1}{49}-\frac{1}{50}\)

\(=\left(1+\frac{1}{3}+\frac{1}{5}+.......+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+........+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+.......+\frac{1}{50}\right)\)

\(=\left(1+\frac{1}{2}+\frac{1}{3}+........+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+......+\frac{1}{25}\right)\)

\(=\frac{1}{26}+\frac{1}{27}+........+\frac{1}{50}\left(đpcm\right)\)

\(4-x^2+2x=-\left(x^2-2x-4\right)=-\left(x^2-2x+1+3\right)\)

\(=-\left[\left(x-1\right)^2+3\right]=-\left(x-1\right)^2-3\le-3\)

Vậy GTLN của bt là -3\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(4x-x^2-5=-\left(x^2-4x+5\right)=-\left(x^2-4x+4+1\right)\)

\(=-\left[\left(x-2\right)^2+1\right]=-\left(x-2\right)^2-1\le-1\)

Vậy GTLN của bt là -1\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

\(4B=1.2.3.4+2.3.4.4+...+\left(n-1\right)n\left(n+1\right).4\)

         \(=1.2.3.4-0.1.2.3+2.3.4.5-1.2.3.4+...+\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)\(-\left[\left(n-2\right)\left(n-1\right)n\left(n+1\right)\right]\)

          = \(\left(n-1\right)n\left(n+1\right)\left(n+2\right)-0.1.2.3=\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\frac{\left(n-1\right)n\left(n+1\right)\left(n+2\right)}{4}\)