\(2018\cdot2018-2017\cdot2019\)

\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)

\(=2018^2-\left(2018^2-1\right)\)

\(=2018^2-2018^2+1\)

\(=1\)

\(2018.2018-2017.2019\)

\(=2018^2-\left(2018-1\right)\left(2018+1\right)\)

  \(=2018^2-\left(2018^2-1\right)\)

  \(=2018^2-2018^2+1\)

  \(=1\)

ta có

\(2018\times2017+18+2000-2018\times2008=2018\times2018+2018-2018\times2008\)

\(=2018\times\left(2017+1-2008\right)=2018\times10=20180\)

\(\frac{2017}{2018}\)x\(\frac{7}{8}\)+\(\frac{2017}{2018}\)x\(\frac{3}{8}\)-\(\frac{2017}{2018}\)x\(\frac{1}{4}\)

= \(\frac{2017}{2018}\)x  (\(\frac{7}{8}\)+\(\frac{3}{8}\)-\(\frac{1}{4}\))

= \(\frac{2017}{2018}\)x  ( \(\frac{10}{8}\)- \(\frac{1}{4}\))

= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{2}{8}\))

= \(\frac{2017}{2018}\)x  1

= \(\frac{2017}{2018}\)

Chúc em học tốt nhé :>

=2017/2018*(7/8+3/8)-2017*1/4

=2017/2018*5/4+2017*-1/4

=2017/2018*(5/4-1/4)

=2017/2018*1

=2017/2018

\(\dfrac{2017}{2016}\) và \(\dfrac{2017}{2018}\)

C1: Đây là 2 phân số cùng mẫu:

Vì 2016 < 2018 nên \(\dfrac{2017}{2016}>\dfrac{2017}{2018}\)

C2: So sánh với 1.

Vì \(\dfrac{2017}{2016}>1>\dfrac{2017}{2018}\) nên \(\dfrac{2017}{2016}>\dfrac{2017}{2018}\)

Ở trên là 2 phương pháp giải thuận tiện nhất.

C1 là 2 phân số cùng tử số mới đúng nhé, ghi nhầm. Nhưng còn cách so sánh thì đúng cả rồi ạ.

\(\frac{2017}{2018}\times2015+\frac{2017}{2018}\times4-\frac{2017}{2018}.\)

\(=\frac{2017}{2018}\times2015+\frac{2017}{2018}\times4-\frac{2017}{2018}\times1\)

\(=\frac{2017}{2018}\times\left(2015+4-1\right)\)

\(=\frac{2017}{2018}\times2018\)

\(=\frac{2017\times2018}{2018\times1}\)

\(=\frac{2017}{1}=2017\)

2018x36+2018+2018x63

=2018x(36+1+63)

=2018x100

=201800

2018 x 36 + 2018 + 2018 x 63

= 2018 x 36 + 2018 x 1 + 2018 x 63

= 2018 x ( 36 + 1 + 63 )

=2018 x 100

= 201800

^HT^

\(2018\times\frac{3}{4}+2018\times\frac{1}{4}-2018\)

\(=2018\times\left(\frac{3}{4}+\frac{1}{4}-1\right)\)

\(=2018\times0=0\)

B-a:tương tự bài A

b,\(\frac{14}{18}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)

\(=\frac{7}{9}\times\frac{8}{5}-\frac{7}{9}\times\frac{3}{5}\)

\(=\frac{7}{9}\times\left(\frac{8}{5}-\frac{3}{5}\right)\)

\(=\frac{7}{9}\times1=\frac{7}{9}\)\

thanks

B-a,\(\frac{3\times125+3\times125}{6\times43+6\times57}=\frac{2\times3\times125}{6\times\left(43+57\right)}\)

\(=\frac{3\times250}{6\times100}=\frac{5}{2\times2}=\frac{5}{4}\)

thanks