Mik lười quá bạn tham khảo câu 3 tại đây nhé:

Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath

\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)

\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)

\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)

\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)

p=\(\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+49\)

=\(\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\frac{50}{50}\)

=\(\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)

=\(50\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)

p=50*S

\(\frac{S}{\text{p}}=\frac{1}{50}\)

s=1,p=50

Ta có: \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}.\)

Mà\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{8.9}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(=1-\frac{1}{9}=\frac{8}{9}\)

\(\Rightarrow S< \frac{8}{9}\)

Và \(S=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)

Mà \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow S>\frac{2}{5}\)

Vậy: \(\frac{2}{5}< S< \frac{8}{9}\)

a) Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{4^2}< \frac{1}{3\cdot4}\)

. . .

\(\frac{1}{100^2}< \frac{1}{99\cdot100}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2^2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{49\cdot50}\right)\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\left(1+1-\frac{1}{50}\right)\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{4}\cdot\frac{99}{50}=\frac{99}{200}< \frac{100}{200}=\frac{1}{2}\left(đpcm\right)\)

b) Ta có :

\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)

\(\Rightarrow1-\frac{1}{4}+1-\frac{1}{9}+...+1-\frac{1}{2500}>48\)

\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< 49\)

Lại có : \(\frac{1}{2^2}< \frac{1}{1\cdot2}\)

\(\frac{1}{3^2}< \frac{1}{2\cdot3}\)

. . .

\(\frac{1}{50^2}< \frac{1}{49\cdot50}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)

\(\Rightarrow\frac{1}{2^2}+...+\frac{1}{50^2}< \frac{49}{50}< 1\)

\(\Rightarrow-\left(\frac{1}{2^2}+...=\frac{1}{50^2}\right)>1\)

\(\Rightarrow49-\left(\frac{1}{2^2}+...+\frac{1}{50^2}\right)>49-1=48\)

hay \(\frac{3}{4}+\frac{8}{9}+...+\frac{2499}{2500}>48\left(đpcm\right)\)

Ta có:\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+....+\frac{48}{2}+\frac{49}{1}+50-50\)

               \(=\left(1+\frac{1}{49}\right)+\left(1+\frac{2}{48}\right)+\left(1+\frac{3}{47}\right)+...+\left(1+\frac{48}{2}\right)+\left(1+\frac{49}{2}\right)-50\)

              \(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+....+\frac{50}{2}+\frac{50}{1}-50\)

              \(=50\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+....+\frac{1}{2}\right)+50-50\)

              \(=50\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+....+\frac{1}{2}\right)\)

mà  \(S=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{49}\)

\(=>\frac{S}{P}=\frac{1}{50}\)

Vậy \(\frac{S}{P}=\frac{1}{50}\)              

( 99 - 1 ) : 2 + 1 = 50 ( số )

làm bừa thui,ai tích mình mình tích lại

Số số hạng là : 

Có số cặp là :

50 : 2 = 25 ( cặp )

Mỗi cặp có giá trị là :

99 - 97 = 2 

Tổng dãy trên là :

25 x 2 = 50

Đáp số : 50

Ta có : \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

       \(\Rightarrow S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

      \(\Rightarrow S>\frac{1}{2}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)           (1)

Ta lại có : \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)

          \(\Rightarrow S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

           \(\Rightarrow S< 1-\frac{1}{9}=\frac{8}{9}\)

Từ (1) và (2)  \(\Rightarrow\frac{2}{5}< S< \frac{8}{9}\)          ( đpcm )

Ta có :

2² = 2.2 < 2.3 => \(\frac{1}{2^2}>\frac{1}{2.3}\)

3² = 3.3 < 3.4 => \(\frac{1}{3^2}>\frac{1}{3.4}\)

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

9² = 9 . 9 < 9. 10 => \(\frac{1}{9^2}>\frac{1}{9.10}\)

=> S > \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

=> S > \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...+\frac{1}{9}-\frac{1}{10}\)

=> S > \(\frac{1}{2}-\frac{1}{10}\)

=> S >\(\frac{2}{5}\)( 1 )

Ta có :

2² = 2 . 2 > 1.2 => \(\frac{1}{2^2}< \frac{1}{1.2}\)

3² = 3.3 > 3.2 => \(\frac{1}{3^2}< \frac{1}{2.3}\)

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

9² = 9.9 > 8.9 => \(\frac{1}{9^2}< \frac{1}{8.9}\)

=> S < \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\)

=> S < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}...+\frac{1}{8}-\frac{1}{9}\)

=> S < \(1-\frac{1}{9}\)

=> S <\(\frac{8}{9}\)( 2 )

Từ ( 1 ) và ( 2 )

=> \(\frac{2}{5}< S< \frac{8}{9}\)

A=19,39033602

làm lần lượt các số hạng rồi sẽ ra