Giúp mình với

ai giải nhanh nhất mình k cho

S=1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81=1-1/81=1/81

80/81 là đúng

vô lí vì 2/2 = 1 mà 8/9 < 1

hình như phân số cuối  phải là 1/324
nếu là 1/324 thì tớ giải nè:
S= 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324
= 1/4.(1+1/2^2+1/3^2+1/4^2+1/5^2+1/6^2+1/7^2+1/8^2+1/9^2)                                                    <1/4.(1+1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9)
= 1/4.(1+1-1/9)
= 1/4.17/9 = 17/36<18/36 = 1/2
=> S= 1/4+1/16+1/36+1/64+1/100+1/144+1/196+1/256+1/324<1/2

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!

dpcm là gì vậy các bồ

dpcm là điều phải chứng minh nha

Ta có :     \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

                 \(\frac{1}{16}< \frac{1}{4}-\frac{4}{8}\)

                \(\frac{1}{36}< \frac{1}{8}-\frac{1}{12}\)

                \(\frac{1}{64}< \frac{1}{12}-\frac{1}{16}\)

                 \(\frac{1}{100}< \frac{1}{16}-\frac{1}{20}\)

                   \(\frac{1}{144}< \frac{1}{20}-\frac{1}{24}\)

                 \(\frac{1}{196}< \frac{1}{24}-\frac{1}{28}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{28}\)

             \(=\frac{1}{2}-\frac{1}{28}< \frac{1}{2}\)

Vậy A<1/12