\(\frac{1999.2001-1}{1998+1999.2000}=\frac{1999.2001-\left(1999-1998\right)}{1998+1999.2000}=\frac{1999.2001-1999+1998}{1998+1999.2000}=\frac{1999.\left(20001-1\right)+1998}{1998+1999.2000}=\frac{1999.2000+1998}{1998+1999.2000}=1\)=> 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

\(\frac{1999.2001-1}{1998.1999.2000}.\frac{7}{5}:\frac{14}{15}\)=\(\frac{1.7.15}{1998.5.14}=\frac{1.1.3}{1998.1.2}=\frac{3}{3996}=\frac{1}{1332}\)

\(A=\frac{1999\times\left(2000+1\right)-1}{1998\times1999\times2000}\times\frac{7}{5}\times\frac{15}{14}=\frac{1999\times2000+1999-1}{1998\times1999\times2000}\times\frac{7}{5}\times\frac{5\times3}{7\times2}\)

\(A=\frac{1999\times2000+1998}{1998\times1999\times2000}\times\frac{3}{2}=\frac{3999998\times3}{3\times666\times1999\times2000\times2}=\frac{1999999\times2}{666\times1999\times2000\times2}=\frac{1999999}{666\times1999\times2000}=...\)

Em xem lại đề: có thể đề là: 

A = \(\frac{1999\times2001-1}{1998+1999\times2000}\times\frac{7}{5}:\frac{14}{15}\)= \(\frac{1999\times2000+1999-1}{1998\times1999\times2000}\times\frac{7}{5}\times\frac{5\times3}{7\times2}\)= \(\frac{1999\times2000+1998}{1998+1999\times2000}\times\frac{3}{2}=1\times\frac{3}{2}=\frac{3}{2}\)

Ta sẽ rút gọn phép tính trên bằng cách:Ta loai số 1999 hàng trên và hàng dưới.Tiếp theo đó ta thực hiện phép tính trừ và bằng:2001-1=2000

Vậy ta gạch bỏ 2000 ở trên và dưới và bằng 1/1998*7/5:14/15=15/19980=3/3996=1/1332

Ttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttttat ca nhan voi 7/5 nua

A=1

A=(3999999-1)/(1998+3998000)=3999998/3999998=1

1999.2000+1999-1/1998+1999.2000

1999.2000+1998/1998+1999.2000

=1

\(A=\frac{1999\times2001-1}{1998+1999\times2000}=\frac{1999\times2000+1999-1}{1998+1999\times2000}=\frac{1999\times2000+1998}{1998+1999\times2000}=1\)

Ko vo A =

\(\frac{1999\cdot2001-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot\left(2000+1\right)-1}{1998+1999\cdot2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot2000+1999-1}{1998+1999.2000}\cdot\frac{7}{5}\)

\(=\frac{1999\cdot2000+1998}{1998+1999.2000}\cdot\frac{7}{5}=1\cdot\frac{7}{5}=\frac{7}{5}\)

thank you very much

1999.2001-1/1998+1999.2000 x 7/5

=1999.2001-(1999-1998)/1998+1999.2000 x7/5

=1999.2001-1999+1998/1998+1999.2000 x7/5

=1999(2001-1)+1998/1998+1999.2000x7/5

=1999.2000+1998/1998+1999.2000x7/5

=1x7/5=7/5

TẦM NHƯ HƠI CĂNG

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

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

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

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

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

1) Có nhận xét sau:

\(\frac{1}{a\sqrt{a+1}+\left(a+1\right)\sqrt{a}}=\frac{1}{\sqrt{a^2+a}\left(\sqrt{a}+\sqrt{a+1}\right)}=\frac{\sqrt{a+1}-\sqrt{a}}{\sqrt{a^2+a}}\)

\(=\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{a+1}}.\)Do đó biểu thức có giá trị bằng: \(\frac{1}{1}-\frac{1}{\sqrt{2}}+..-\frac{1}{\sqrt{1999}}=1-\frac{1}{\sqrt{1999}}.\)

2) Có nhận xét sau:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a+1}-\sqrt{a}}{\left(\sqrt{a}+\sqrt{a+1}\right)\left(\sqrt{a+1}-\sqrt{a}\right)}=\sqrt{a+1}-\sqrt{a}.\) Thay vào biểu thức ta được biểu thức

có giá trị bằng: \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{1999}-\sqrt{1998}=\sqrt{1999}-1.\)

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

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

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

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

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

\(=-555,8\)

Hình như sai đề

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

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

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

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

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

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

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

\(=400\)

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

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

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

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