A=1.3/2^2.2.4/3^2.3.5/4^2...99.101/100.100

A=(1.2.3...99/2.3.4...100).(3.4.5...101/2.3.4...100)

A=1/100.101/2

A=101/200

200.A=200.101/200

200.A=101

phan h cac so tren ra roi nhan lai la dc xin loi nha may binh bi liet nen ko giai dc

\(P=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.....\frac{10000}{9999}=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{100.100}{99.101}\)

\(=\frac{\left(2.3....100\right).\left(2.3.4...100\right)}{\left(1.2.3.4...99\right).\left(3.4.....101\right)}=\frac{100.2}{101}=\frac{200}{101}\)

算:4/3*9/8*16/15*25/24*.*9801/989992014-12-1+2+3+4+5+6+7+8+9+10.+9999+100=（）

9999+999+99+9 9999×2222+3333×3334 10000-1-2-3-4-...-80 1+2+3+4+...+78+79+80 2007*2005/2000 要简算

1+2+3+4+5+6+.+9999+10000=

10000-9999+9998-9997…+6-5+4-3+2-1等于多少

=(2/1*3)*(3平方/2*4)*(4/3*5)*(5平方/4*6)................*(99平方/98*100)*(100平方/99*101)
=2*100/101=200/101

Ta có :

\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(A=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)

\(A=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)\)

\(A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99\)\(\left(1\right)\)

gọi B là biểu thức trong ngoặc

Lại có :

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

\(B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(B< 1-\frac{1}{100}< 1\)

\(\Rightarrow A=99-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>99-\left(1-\frac{1}{100}\right)>98\)

\(\Rightarrow A>98\)\(\left(2\right)\)

từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(98< A< 99\)

vậy A không phải là số tự nhiên

phần bạn đánh dấu (1) thì A<99 vì A= 99 trừ đi một số mà

\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)

\(\frac{3}{4}\)*\(\frac{8}{9}\)*\(\frac{15}{16}\)********\(\frac{9999}{10000}\)

= \(\frac{1\cdot3}{2^2}\)*\(\frac{2\cdot4}{3^2}\)********\(\frac{99\cdot101}{100^2}\)

= \(\frac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\)* \(\frac{3\cdot4\cdot5\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot100}\)

= \(\frac{1}{100}\)*\(\frac{101}{2}\)=\(\frac{101}{200}\)

Ta có: A = \(\frac{3}{8}\). \(\frac{8}{9}\).\(\frac{15}{16}\). ... .\(\frac{9999}{10000}\)
\(\Rightarrow\) A = \(\frac{1.3}{2^2}\).\(\frac{2.4}{3^2}\). \(\frac{3.5}{4^2}\). ... . \(\frac{99.101}{100^2}\)
\(\Rightarrow\) A = \(\frac{1.111}{2.100}\)= \(\frac{111}{200}\)
Vậy: A = \(\frac{111}{200}\).

\(A=\frac{\left(1\cdot3\right)\cdot\left(2\cdot4\right)\cdot\left(3\cdot5\right)\cdot.....\cdot\left(99\cdot101\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right).......\left(100\cdot100\right)}\)

\(A=\frac{\left(1\cdot2\cdot3\cdot.....\cdot99\right)\left(3\cdot4\cdot5\cdot.....\cdot101\right)}{\left(2\cdot3\cdot4\cdot.....\cdot100\right)\left(2\cdot3\cdot4\cdot.....\cdot100\right)}\)

\(A=\frac{1\cdot101}{100\cdot2}\)

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

k mk nha mk nhanh nhất mk trả lời đúng 100%

\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)

\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)

\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)

\(x=\frac{1}{100}.\frac{101}{2}\)

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

\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)

\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)

\(X=\frac{1}{100}.\frac{101}{2}\)

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

Study well 

~ So sad :( !! ~

\(A=\frac{31}{60}\)

I thinks so ! Sad