1/2-1/3+1/4-1/5=13/60>12/60=0,2

tiếp tục gom vd 1/6>1/7=>1/6-1/7>0

cứ như thế

A>0,2

tương tự như trên ha!

OK. Tối nhớ giải hộ mik nha

Mik hứa sẽ lik-e cho bạn

mình ko biết

\(\left(1\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...+\left(\frac{1}{98}-\frac{1}{99}\right)\)

\(=\frac{13}{60}+\left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+...\left(\frac{1}{98}-\frac{1}{99}\right)\)

Ta thấy        \(\frac{13}{60}>\frac{12}{60}=\frac{1}{5}\)

          \(\frac{1}{6}-\frac{1}{7}>0\)

          \(\frac{1}{8}-\frac{1}{9}>0\)

\(...\)\(\frac{1}{98}-\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)

\(\left(2\right)\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

\(=\frac{23}{60}-\left(\frac{1}{7}-\frac{1}{8}\right)-\left(\frac{1}{9}-\frac{1}{10}\right)-...-\left(\frac{1}{97}-\frac{1}{98}\right)-\frac{1}{99}\)

Ta thấy   \(\frac{23}{60}< \frac{24}{60}=\frac{2}{5}\)

      \(\frac{1}{7}-\frac{1}{8}>0\)

     \(\frac{1}{9}-\frac{1}{10}>0\)

\(...\frac{1}{97}-\frac{1}{98}>0\)

                \(\frac{1}{99}>0\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)

Đặt \(B=1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\)

\(=\left(1+\frac{1}{99}\right)+\left(\frac{1}{3}+\frac{1}{97}\right)+\left(\frac{1}{5}+\frac{1}{95}\right)+...+\left(\frac{1}{49}+\frac{1}{51}\right)\)

\(=\frac{100}{99}+\frac{100}{3\times97}+\frac{100}{5\times95}+...+\frac{100}{49\times51}\)

\(=100\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

Đặt \(C=\frac{1}{1\times99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{97\times3}+\frac{1}{99\times1}\)

\(=2\left(\frac{1}{99}+\frac{1}{3\times97}+\frac{1}{5\times95}+...+\frac{1}{49\times51}\right)\)

\(A=\frac{B}{6}=\frac{100}{2}=50\)

Vậy \(A=50\)

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khó quá bạn ơi!

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

\(A=\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{98}\right)-\left(\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)\Rightarrow-A+1=\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{98}\right)=\left(1+\frac{1}{2}+....+\frac{1}{99}\right)-2\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{98}\right)=\left(1+\frac{1}{2}+...+\frac{1}{99}\right)-\left(1+\frac{1}{2}+......+\frac{1}{49}\right)=\frac{1}{50}+....+\frac{1}{99}\)

\(\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{99}=\left(\frac{1}{50}+\frac{1}{51}+....+\frac{1}{69}\right)+\left(\frac{1}{70}+\frac{1}{71}+.....+\frac{1}{79}\right)+\left(\frac{1}{80}+....+\frac{1}{99}\right)< 20.\frac{1}{50}+10.\frac{1}{70}+20.\frac{1}{80}=\frac{2}{5}+\frac{1}{7}+\frac{1}{4}=\frac{111}{140}< \frac{112}{140}=\frac{4}{5}\Rightarrow-A+1< \frac{4}{5}\Leftrightarrow-A< -0,2\Leftrightarrow A>0,2\)

\(\frac{1}{50}+\frac{1}{51}+....+\frac{1}{99}=\left(\frac{1}{50}+\frac{1}{51}+.....+\frac{1}{59}\right)+\left(\frac{1}{60}+\frac{1}{61}+.....+\frac{1}{69}\right)+....+\left(\frac{1}{90}+\frac{1}{91}+...+\frac{1}{99}\right)>\frac{1}{6}+\frac{1}{7}+...+\frac{1}{9}+\frac{1}{10}>\frac{2}{7}+\frac{2}{9}+\frac{1}{10}=\frac{383}{630}>\frac{378}{630}=\frac{3}{5}\Rightarrow-A+1>\frac{3}{5}\Leftrightarrow-A>-\frac{2}{5}\Leftrightarrow A< \frac{2}{5}=0,4\Rightarrow0,2< A< 0,4\)