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

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

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

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

      \(\frac{1}{7}-\frac{1}{8}>0\)

     \(\frac{1}{9}-\frac{1}{10}>0\)

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

                \(\frac{1}{99}>0\)

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

Ta có :

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

biểu thức trong dấu ngoặc thứ nhất bằng \(\frac{13}{60}\)nên lớn hơn \(\frac{12}{60}\), tức là lớn hơn 0,2, còn các dấu ngoặc sau đều dương, do đó :

A > 0,2

để chứng minh A < 0,4 hay \(\frac{2}{5}\)

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

biểu thức trong dấu ngoặc thứ nhất nhở hơn \(\frac{2}{5}\), còn các dấu ngoặc sau đều dương,

do đó A < \(\frac{2}{5}\)hay A < 0,4

Vậy 0,2 < A < 0,4

Bài này giải ra dài lắm.

mình đang cần bài này

1<B<2 đi cho lẹ

ko can phai tinh ma la phan tich a vua co 1/15 vua co 1/10 thi co the so sanh ok?

Thế thì phải tính A à

Vậy bạn tính đi

Đặt : \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)

Ta thấy :

\(\frac{1}{5^2}< \frac{1}{4.5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(\frac{1}{7^2}< \frac{1}{6.7}\)

\(.......................\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}\)

Vì \(\frac{1}{6}< \frac{6}{25}< \frac{1}{4}\)nên \(\frac{1}{6}< A< \frac{1}{4}\)hay \(\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\)

~ Hok tốt ~

Bài 1:

Đặt  \(A=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)

Ta có: 

\(A< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)

Ta có:

\(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}>\frac{1}{6}\)

\(\Rightarrow\frac{1}{6}< \frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(\text{đ}pcm\right)\)

Bài 2:

\(a)\)Tách tổng A thành ba nhóm:

\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{70}\right)\)

\(A>\frac{1}{30}\cdot20+\frac{1}{50}\cdot20+\frac{1}{70}\cdot20=\frac{2}{3}+\frac{2}{5}+\frac{2}{7}=1\frac{37}{105}\)

\(A>1\frac{35}{105}=1\frac{1}{3}=\frac{4}{3}\left(\text{đ}pcm\right)\)

\(b)\)Tách tổng A thành sáu nhóm:

\(A=\left(\frac{1}{11}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+...+\frac{1}{50}\right)\)\(+\left(\frac{1}{51}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+...+\frac{1}{70}\right)\)

\(A< \frac{1}{11}\cdot10+\frac{1}{21}\cdot10+\frac{1}{31}\cdot10+\frac{1}{41}\cdot10+\frac{1}{51}\cdot10+\frac{1}{61}\cdot10\)

\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{4}+\frac{1}{5}\right)< 2+0,5=2,5\left(\text{đ}pcm\right)\)

#Sakura

Bài 1:

ta thấy \(\frac{1}{5^2}< \frac{1}{4.5};\)

            \(\frac{1}{6^2}< \frac{1}{5.6}\)

.................

=>1/_5²  +1/_6²+.....+1/_100²<1/(4.5)+1/(5.6)+.....+1/(99+100)

=>1/_5²  +1/_6²+.....+1/_100² <1/4 -1/5 +1/5 -1/6 +......+1/99 -1/100

=>1/_5²  +1/_6²+.....+1/_100² <1/4 - 1/100 <1/4

CMTT với 1/_5² >1/(5.6).......

=>1/_5² +1/_6²+.....+1/_100² >1/5 -1/6 +1/6 - 1/7+ .........+1/100 - 1/101

=>1/_5² +1/_6²+.....+1/_100² >1/5 - 1/101>1/5>1/6