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

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

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

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

= 1 \(\times\) \(\dfrac{7}{5}\)

= \(\dfrac{7}{5}\)

trời đát cẩn thạn

bị trừ điểm đó bạn ạ vì câu hỏi thế này

thì quá dài và ....

\(\frac{1999\times\left(2000+1\right)-1}{1998+1999\times2000}\times\frac{7}{5}\Rightarrow\frac{1999\times2000+1999\times1-1}{1998+1999\times2000}\times\frac{7}{5}=\frac{1999\times2000+1998}{1998+1999\times2000}\times\frac{7}{5}=1\Rightarrow1\times\frac{7}{5}=\frac{7}{5}\)

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

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

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

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

\(=1.\)

Đúng 100%

bằng 1 nhé

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

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

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

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

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

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

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

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