đáp số cuối cùng là 0 vì 1-1/4 = 0/4 = 0 x cho những số nào khác cũng bằng 0 thôi

\(\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times\left(1-\frac{1}{16}\right)\times...\times\left(1-\frac{1}{10000}\right)\)

\(=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{999}{10000}\)

\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times...\times\frac{99\times101}{100\times100}\)

\(=\frac{1}{2}\times\frac{101}{100}\)

\(=\frac{101}{200}\)

Chu Nguyen Nhan sai rồi nhé 1-1/4 mà bằng 0 à        Học lại đê

\(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right).....\left(1-\dfrac{1}{10000}\right)\)

\(=\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot\dfrac{4^2-1}{4^2}\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)

\(=\dfrac{1.3.2.4.3.5.....99.101}{2.2.3.3.4.4....100.100}\)

\(=\dfrac{\left(1.2.3...99\right)}{2.3.4....100}\cdot\dfrac{3.4.5...101}{2.3.4....100}=\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{101}{200}\)

Ta có: \(\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{10000}\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-9999}{10000}\)

\(=-\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\)

\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot1111\cdot9}{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}\)

\(=\dfrac{101}{200}\)

\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\cdot...\cdot\left(1-\dfrac{1}{9801}\right)\)

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\left(1-\dfrac{1}{99}\right)\left(1+\dfrac{1}{99}\right)\)

\(=\left(\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{98}{99}\right)\cdot\left(\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{100}{99}\right)\)

\(=\dfrac{1}{99}\cdot\dfrac{100}{2}=\dfrac{50}{99}\)

A=1.2.3+2.3.4+....+99.100.101

4A=1.2.3.4+2.3.4.(5-1)+3.4.5.(6-2)+....+98.99.100.(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-3.4.5.2+....+98.99.100.101-98.99.100.97

4A=98.99.100.101

4A=97990200

A=97990200/4

A=24497550

B=1.2+3.4+5.6+7.8+8.9+...+999.1000

3B=1.2.3+2.3.(4-1)+3.4(5-2)+....+998.999(1001-998)

3B=1.2.3+2.3.4-2.3.1+3.4.5-3.4.2+....+998.999.1001-998.999.998

3B=999.1000.1001

3B=999999000

B=999999000/3

B=333333000

C=1+4+9+16+25+36+.....+10000

C=1^2+2^2+3^2+4^2+5^2+6^2+....+100^2

C=(1^2+3^2+5^2+.....+99^2)+(2^2+4^2+6^2+....+100^2)

C=99.100.101/6  +   100.101.102/6

C=166650         +171700

C=338350

Còn câu d bạn  dựa vào câu c là làm được ngay bây h mk mỏi tay rùi ko muốn đánh nữa khi nào rảnh mk gửi công thức cho nha bây h mk bận rùi.

chúc bn học tốt

   A=1.2.3+2.3.4+....+99.100.101

4.A=1.2.3.(4-0)+2.3.4.(5-1)+...+99.100.101.(102-98)

4.A=1.2.3.1-0.1.2.3+2.3.4.5-1.2.3.4+....+99.100.101.102-98.99.100.101

4.A=99.100.101.102

  A=\(\frac{99.100.101.102}{4}\)

   B=1.2+2.3+3.4+...+999.1000

3.B=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+.....+999.1000.(1001-998)

3.B=1.2.3-0.1.2+2.3.4-1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+......+999.1000.1001-998.999.1000

3.B=999.1000.1001

=>B=\(\frac{999.1000.1001}{3}\)

C và D dễ lắm bạn tự làm nhé

mình lại có cách làm khác là tìm số thừa ra lấy số cuối cộng số đầu chia cho  2 . tính số các số hạng là số cuối trừ số đầu chia khoảng cách cộng 1  , roi tinh gia tri 1 cap la so cuoi cong so dau sau do tinh so cap la lay so cac so hang chia 2 va di tinh tong thi ban tu lam

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

1) 3/9 + 5/9 = 8/9

2) 1/2 x 2/3 x 3/4 x 4/5

= 1/5

(1-1/2)×(1-1/3)×(1-1/4)×(1-1/5)×1-1/6=1/2×2/3×3/4×4/5×5/6=1/6

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

x = nhân ạ

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)

\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)

\(=\dfrac{1}{5}\)

TÍNH BẰNG CÁCH NHANH NHẤT NHA CÁC BN 

a) \(\left(\frac{1}{3}+\frac{1}{5}\right)+\left(\frac{1}{6}-\frac{1}{5}\right)=\left(\frac{1}{3}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)=\frac{1}{2}\)

b) \(\frac{3}{16}\times\frac{7}{5}+\frac{3}{5}\times\frac{9}{16}=\frac{21}{80}+\frac{27}{80}=\frac{48}{80}=\frac{3}{5}\)

c) \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2020\times2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2020}-\frac{1}{2021}\)

\(=1-\frac{1}{2021}=\frac{2020}{2021}\)

d) \(\frac{1}{1\times3}+\frac{1}{3\times5}+...+\frac{1}{2021\times2023}=\frac{1}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{2021\times2023}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2021}-\frac{1}{2023}\right)\)

\(=\frac{1}{2}\times\left(1-\frac{1}{2023}\right)=\frac{1}{2}\times\frac{2022}{2023}=\frac{1011}{2023}\)

e) \(\frac{3}{2}\times\frac{1}{7}\times\frac{5}{4}+\frac{15}{2}\times\frac{6}{7}\times\frac{1}{4}==\frac{15}{56}+\frac{80}{56}=\frac{95}{56}\)