\(=\frac{199.2000+199-1}{1998+1999.2000}.\frac{7}{5}\)

\(=\frac{199.2-1}{1998-1999}.\frac{7}{5}\)

\(=\frac{398-1}{-1}.\frac{7}{5}\)

\(=\frac{397}{-1}.\frac{7}{5}\)

\(=-397.\frac{7}{5}\)

\(=-555,8\)

Hình như sai đề

\(\frac{1999.2001-1}{1998+1999.2000}=\frac{1999.2001-\left(1999-1998\right)}{1998+1999.2000}=\frac{1999.2001-1999+1998}{1998+1999.2000}=\frac{1999.\left(20001-1\right)+1998}{1998+1999.2000}=\frac{1999.2000+1998}{1998+1999.2000}=1\)=> đáp án là 7/5

có bị thiếu dấu ngoặc không vậy

1999 x 2001 - 1 =3999998 ; 1998+1999 x 2000 =3999998 

Suy ra : \(\frac{3999998}{3999998}\)=1 Suy ra 1x7/5 = 7/5

\(1\frac{1}{5}\cdot1\frac{1}{6}\cdot1\frac{1}{7}\cdot...\cdot1\frac{1}{1998}\cdot1\frac{1}{1999}\)

\(=\frac{6}{5}\cdot\frac{7}{6}\cdot\frac{8}{7}\cdot...\cdot\frac{1999}{1998}\cdot\frac{2000}{1999}\)

\(=\frac{6\cdot7\cdot8\cdot...\cdot1999\cdot2000}{5\cdot6\cdot7\cdot...\cdot1998\cdot1999}\)

\(=\frac{2000}{5}=400\)

\(1\frac{1}{5}‧1\frac{1}{6}‧1\frac{1}{7}‧.......‧1\frac{1}{1998}‧1\frac{1}{1999}\)

\(=\frac{6}{5}‧\frac{7}{6}\frac{8}{7}‧.......‧\frac{1999}{1998}‧\frac{2000}{1999}\)

\(=\frac{6‧7‧8‧.......‧1999‧2000}{5‧6‧7‧.......‧1998‧1999}\)

\(=400\)

\(1\frac{1}{5}\cdot1\frac{1}{6}\cdot1\frac{1}{7}\cdot...\cdot1\frac{1}{1998}\cdot1\frac{1}{1999}\)

\(=\frac{6}{5}\cdot\frac{7}{6}\cdot\frac{8}{7}\cdot...\cdot\frac{1999}{1998}\cdot\frac{2000}{1999}\)

\(=\frac{6\cdot7\cdot8\cdot...8\cdot1999\cdot2000}{5\cdot6\cdot7\cdot...\cdot1998\cdot1999}\)

\(=\frac{2000}{5}=400\)

Sửa lại :

\(=\frac{1999.2000+1999-1}{1998+1999.2000}.\frac{7}{5}=\frac{1999.2000+1998}{1998+1999.2000}.\frac{7}{5}=1.\frac{7}{5}=\frac{7}{5}\)

\(\frac{1999.2000-1}{1998+1999.2000}.\frac{7}{5}\)

= \(\frac{-1}{1998}\) . \(\frac{7}{5}\)

= \(\frac{-7}{9990}\)

\(\frac{1999.2000+1999}{1998+1999.2000}.\frac{7}{5}=1+\frac{1}{1998+1999.2000}.\frac{7}{5}=1+\frac{7}{5.\left(1998+1999.2000\right)}=1+\frac{7}{19999990}=\frac{19999997}{19999990}\)

b) \(\frac{1}{1000}+\frac{13}{1000}+\frac{25}{1000}+...+\frac{87}{1000}+\frac{99}{1000}\)

\(=\frac{1+13+25+...+85+97}{1000}=\frac{\left(97+1\right).\left[\left(97-1\right):12+1\right]:2}{1000}\)

\(=\frac{49.9}{1000}=\frac{441}{1000}.\) ( Đề bài sai nhé bạn tử số : 1; 13; 25; 37; 49 ; 61; 73; 85 ; 97. )

giai cho tao nhanh ko con bao

TẦM NHƯ HƠI CĂNG

\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{1999}{1}+\frac{1998}{2}+\frac{1997}{3}+....+\frac{1}{1999}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{2000}}{1+\left(\frac{1998}{2}+1\right)+\left(\frac{1997}{3}+1\right)+....+\left(\frac{1}{1999}+1\right)}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{\frac{2000}{2}+\frac{2000}{3}+\frac{2000}{4}+....+\frac{2000}{2000}}\)

\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}{2000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}\)

\(=\frac{1}{2000}\)