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 =

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

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

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

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

ta có 

\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)

ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)

\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)

hay \(S_3=0+0+..+0=0\)

\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)

\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)

\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)

\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)

\(=1\)

Bạn chỉ giùm mình từng bước nha, cảm ơn

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

\(=\frac{\left[\frac{2001}{1}+1\right]+\left[\frac{2001}{2}+1\right]+...+\left[\frac{2001}{2000}+1\right]+2001}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}\)

\(=\frac{2001\left[1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}\right]}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2000}}=2001\)

$\ge $ 

là sao?

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