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#)Giải :
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{2499}{2500}\)
\(A=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times\frac{4.6}{5.5}\times...\times\frac{49.51}{50.50}\)
\(A=\frac{1\times3\times2\times4\times3\times5\times...\times49\times51}{2\times2\times3\times3\times4\times4\times...\times50\times50}\)
\(A=\frac{1\times51}{2\times50}\)
\(A=\frac{51}{100}\)
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{2499}{2500}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\frac{6\times4}{5\times5}\times...\times\frac{49.51}{50\times50}\)
\(=\frac{1}{2}\times\frac{51}{50}\)
\(=\frac{51}{100}\)

\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}\)
\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{29.31}{30.30}\)
\(A=\dfrac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4.....30.30}\)
\(A=\dfrac{1.2.3.....29}{2.3.4....30}.\dfrac{3.4.5.....31}{2.3.4.....30}\)
\(A=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
\(B=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}\)
\(B=\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}.....\dfrac{49.51}{50.50}\)
\(B=\dfrac{2.4.3.5.4.6.....49.51}{3.3.4.4.5.5....50.50}\)
\(B=\dfrac{2.3.4......49}{3.4.5....50}.\dfrac{4.5.6.....51}{3.4.5....50}\)
\(B=\dfrac{2}{50}.\dfrac{51}{3}=\dfrac{17}{25}\)
Giải:
\(A=\dfrac{3}{2^2}.\dfrac{8}{3^2}.\dfrac{15}{4^2}.....\dfrac{899}{30^2}.\)
\(A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{29.31}{30^2}.\)
\(A=\dfrac{1.2.3.....29}{2.3.4.....30}.\dfrac{2.3.4.....31}{2.3.4.....30}.\)
\(A=\dfrac{1}{30}.31=\dfrac{30}{31}.\)
Vậy \(A=\dfrac{30}{31}.\)

S=43+98+...+25002499
\(= \frac{2^{2} - 1}{2^{2}} + \frac{3^{2} - 1}{3^{2}} + . . . + \frac{5 0^{2} - 1}{5 0^{2}}\)
\(= \left(\right. 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)
\(= 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)
\(\frac{1}{2^{2}} < \frac{1}{1 \cdot 2} = 1 - \frac{1}{2}\)
\(\frac{1}{3^{2}} < \frac{1}{2 \cdot 3} = \frac{1}{2} - \frac{1}{3}\)
...
\(\frac{1}{5 0^{2}} < \frac{1}{49 \cdot 50} = \frac{1}{49} - \frac{1}{50}\)
Do đó: \(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + . . . + \frac{1}{49} - \frac{1}{50} = 1 - \frac{1}{50}\)
=>\(\frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1\)
=>\(0 < \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} < 1\)
=>\(0 > - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > - 1\)
=>\(0 + 49 > - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) + 49 > - 1 + 49\)
=>49>B>48
=>S không là số tự nhiên
S=43+98+1615+...+50002499
\(S = 1 - \frac{1}{4} + 1 - \frac{1}{9} + 1 - \frac{1}{16} + . . . + 1 - \frac{1}{5000}\)
\(S = \left(\right. 1 + 1 + 1 + . . . + 1 \left.\right) - \left(\right. \frac{1}{4} + + \frac{1}{9} + \frac{1}{16} + . . . + \frac{1}{5000} \left.\right)\)
\(S = 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) < 49\)\(\left(\right. 1 \left.\right)\)
Lại có :
\(\frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} < \frac{1}{1.2} + \frac{1}{2.3} + \frac{1}{3.4} + . . . + \frac{1}{49.50}\)
\(= \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . . + \frac{1}{49} - \frac{1}{50} = 1 - \frac{1}{50} < 1\)
\(\Rightarrow\)\(- \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > - 1\)
\(\Rightarrow\)\(S = 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) > 49 - 1 = 48\)\(\left(\right. 2 \left.\right)\)
Từ (1) và (2) suy ra :
\(48 < S < 49\)
Vậy S không là số tự nhiên
Chúc các bạn học tốt nhé ! =))

a=8/9+15/16+24/25+....+2499/2500
a=(1-1/9)+(1-1/16)+(1-1/25)+....+(1-1/2500)
a=1-1/9+1-1/16+1-1/25+....+1-1/2500
a=(1+1+...+1)-(1/9+1/16+1/25+....+1/2500)
Tính \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.............................\dfrac{2499}{2500}\)

A=2.4/3^2 . 3.5/4^2 . 4.6/5^2 ............ . 49.51/50^2
A=2/3-51/50
A=17/25.
Chúc bạn hok tốt.
Bài này cũng dễ ý mà, vô cùng đơn giản.........
Giải:
Ta có: \(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}.....\dfrac{2499}{2500}.\)
\(=\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}.....\dfrac{49.51}{50^2}.\)
\(=\dfrac{\left(2.3.4.....49\right)\left(4.5.6.....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}.\)
\(=\dfrac{2.51}{3.50}.\)
\(=\dfrac{17}{25}.\)
CHÚC BN HỌC TỐT!!! ^ _ ^
Đừng quên bình luận nếu bài mik sai nhé!!! - _ -
Còn nếu bài mik đúng thì nhớ tick mik để mik lấy SP nha!!! ^ - ^

A= 3^2-1/3.3 . 4^2-1/4.4 . 5^2-1/5.5 . ... 50^2-1/50.50 A= (3+1).(3-1).(4+1).(4-1).(5+1).(5-1). ... (50+1).(50-1) / 3.3.4.4.5.5. ... . 50.50 A=4.2.5.3.6.4. ... 51.49 / 3.3.4.4.5.5....50.50 A=(4.5.6. ... .51).(2.3.4. ... 49)/(3.4.5.... .50).(3.4.5.. ... 50) A= 51.2/3.50 A=17/25
Ta có:
\(A=\dfrac{8}{9}.\dfrac{15}{16}.\dfrac{24}{25}......\dfrac{2499}{2500}\)
= \(\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.\dfrac{4.6}{5.5}......\dfrac{49.51}{50.50}\)
= \(\dfrac{2.4.3.5.4.6......49.51}{3.3.4.4.5.5......50.50}\)
= \(\dfrac{\left(2.3.4....49\right)\left(4.5.6....51\right)}{\left(3.4.5....50\right)\left(3.4.5....50\right)}\)
= \(\dfrac{2}{50}.\dfrac{51}{3}\) = \(\dfrac{17}{25}\)

\(\Leftrightarrow A=\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4.4}...\frac{49.51}{50.50}\)
\(\Rightarrow A=\frac{2\cdot4\cdot3\cdot5\cdot...\cdot49.51}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(\Rightarrow A=\frac{2\cdot51}{3\cdot50}\)
\(\Rightarrow A=\frac{17}{25}\)

Bạn vào https://sites.google.com/site/toantieuhocpl/20-tinh-nhanh sẽ có đấy.
=2.4/3^2.3.5/4^2.4.6/5^2.....49.51/50^2
=(2.3.4.....49).(4.5.6.....51)/(3.4.5.....50).(3.4.5.....50)
=2.51/50.3
=17/25
câu hỏi đâu bn?
Nếu câu hỏi là Chứng tỏ B không phải là số nguyên thì:
B \(= \frac{2^{2} - 1}{2^{2}} + \frac{3^{2} - 1}{3^{2}} + \frac{4^{2} - 1}{4^{2}} + . . . + \frac{5 0^{2} - 1}{5 0^{2}}\)
\(= 49 - \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right)\)
mà \(0 < \left(\right. \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4^{2}} + . . . + \frac{1}{5 0^{2}} \left.\right) < 1\)cũng như \(\notin Z\)
Vậy B không phải là số nguyên ^_^