Áp dụng công thức k/n*m=k/n-k/m trong đó n-m=k hoặc m-n=k
thế vào ta có
A=1/2*3+1/4*5+...+1/98*99
tớ biết tới đó thôi để từ từ tớ suy nghĩ rồi trả lời cho
\(A=\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{98}\right)-\left(\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)\Rightarrow-A+1=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{98}\right)=\left(1+\frac{1}{2}+....+\frac{1}{99}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{98}\right)=\left(1+\frac{1}{2}+...+\frac{1}{99}\right)-\left(1+\frac{1}{2}+......+\frac{1}{49}\right)=\frac{1}{50}+....+\frac{1}{99}\)
\(\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{99}=\left(\frac{1}{50}+\frac{1}{51}+....+\frac{1}{69}\right)+\left(\frac{1}{70}+\frac{1}{71}+.....+\frac{1}{79}\right)+\left(\frac{1}{80}+....+\frac{1}{99}\right)< 20.\frac{1}{50}+10.\frac{1}{70}+20.\frac{1}{80}=\frac{2}{5}+\frac{1}{7}+\frac{1}{4}=\frac{111}{140}< \frac{112}{140}=\frac{4}{5}\Rightarrow-A+1< \frac{4}{5}\Leftrightarrow-A< -0,2\Leftrightarrow A>0,2\)
\(\frac{1}{50}+\frac{1}{51}+....+\frac{1}{99}=\left(\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{59}\right)+\left(\frac{1}{60}+\frac{1}{61}+.....+\frac{1}{69}\right)+....+\left(\frac{1}{90}+\frac{1}{91}+...+\frac{1}{99}\right)>\frac{1}{6}+\frac{1}{7}+...+\frac{1}{9}+\frac{1}{10}>\frac{2}{7}+\frac{2}{9}+\frac{1}{10}=\frac{383}{630}>\frac{378}{630}=\frac{3}{5}\Rightarrow-A+1>\frac{3}{5}\Leftrightarrow-A>-\frac{2}{5}\Leftrightarrow A< \frac{2}{5}=0,4\Rightarrow0,2< A< 0,4\)
