\(B=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{10.11}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{2}-\dfrac{1}{11}< \dfrac{1}{2}\)
Bài 6:
\(A=\left(x+\dfrac{1}{2}\right)^2+\left|2y-\dfrac{3}{4}\right|+\dfrac{175}{3}\ge\dfrac{175}{3}\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(-\dfrac{1}{2};\dfrac{3}{8}\right)\)