\(\frac{1999x2001-1}{1998+1999x2000}x\frac{7}{5}=\frac{1999x\left(2000+1\right)-1}{1998+1999x2000}x\frac{7}{5}\)

\(=\frac{1999x2000+1999-1}{1998+1999x2000}x\frac{7}{5}\)

\(=\frac{1999x2000+1998}{1998+1999x2000}x\frac{7}{5}\)

\(=1x\frac{7}{5}\)

\(=\frac{7}{5}\)

\(=\frac{1999\times\left(2000+1\right)-1}{1998+1999\times2000}\times\frac{7}{5}\)

\(=\frac{1999\times2000+1999-1}{1999\times2000+1998}\times\frac{7}{5}\)

\(=1\times\frac{7}{5}=\frac{7}{5}\)

     \(\dfrac{1999\times2001-1}{1998+1999\times2000}\) \(\times\) \(\dfrac{7}{5}\)

= \(\dfrac{1999\times\left(2000+1\right)-1}{1998+1999\times2000}\) \(\times\) \(\dfrac{7}{5}\)

= \(\dfrac{1999\times2000+1999-1}{1998+1999\times2000}\) \(\times\) \(\dfrac{7}{5}\)

= \(\dfrac{1999\times2000+1998}{1999\times2000+1998}\) \(\times\) \(\dfrac{7}{5}\)

= 1 \(\times\) \(\dfrac{7}{5}\)

= \(\dfrac{7}{5}\)

\(=\frac{3999999-1}{1998+3998000}=\frac{3999998}{3999998}=1\)

\(\frac{1999x2001-1}{1998+1999x2000}\)

\(=\frac{1999x2000+1999-1}{1998+1999x2000}\)

\(=\frac{1999x2000+1998}{1998+1999x2000}\)

\(=1.\)

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bằng 1 nhé

a)2007/2008

b)2009x2009/2008x2008

bạn giải chi tiết ra nhé giúp mk với

 1999 x (2000+1)-1=1999x2000+1999-1=1999x2000+1998=1

1998+1999x2000=giữ nguyên=1998+1999x2000=1 Vậy kết quả phân số = 1 k cho mình nhé

\(\frac{1999\times2001-1}{1999+1999\times2000}\times\frac{7}{5}\)

\(=\frac{2001-1}{1999\times2000}\times\frac{7}{5}\)

\(=\frac{2000}{1999\times2000}\times\frac{7}{5}\)

\(=1999\times\frac{7}{5}\)

........... còn lại tự xử lý nha

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