A=1/22+1/32+...+1/92
Ta có:1/22>1/2.3,1/32>1/3.4,...,1/92>1/9.10
⇒A>1/2.3+1/3.4+...+1/9.10
A>1/2-1/3+1/3-1/4+...+1/9-1/10
A>1/2-1/10
A>2/5(đpcm)
Ta có: A = 1/4 + 1/9 + 1/16 + 1/25 +1/36 + 1/49 + 1/64 + 1/81
Vì 1/22>1/2.3,1/32>1/3.4,...,1/92>1/9.10
=>A>1/2.3+1/3.4+...+1/9.10
=>A>1/2-1/3+1/3-1/4+...+1/9-1/10
=>A>1/2-1/10
=>A>2/5
Giải:
\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}+\dfrac{1}{81}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} +\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)
Ta có:
\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\)
\(\dfrac{1}{5^2}=\dfrac{1}{5.5}>\dfrac{1}{5.6}\)
\(\dfrac{1}{6^2}=\dfrac{1}{6.6}>\dfrac{1}{6.7}\)
\(\dfrac{1}{7^2}=\dfrac{1}{7.7}>\dfrac{1}{7.8}\)
\(\dfrac{1}{8^2}=\dfrac{1}{8.8}>\dfrac{1}{8.9}\)
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow A>\dfrac{2}{5}\left(đpcm\right)\)
Chúc bạn học tốt!
