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a: 7/8<8/9<9/10<10/11<11/12
b: 4/5<5/6<6/7<7/8
c: 1/3<3/5<7/9<9/11<11/13
( 1 + 5 + 7 +...+ 2019)\(\times\) ( 4545\(\times\) 51 - 5151 \(\times\) 45) : 51
= ( 1 + 5 + 7 +...+ 2019) \(\times\) ( 45 \(\times\) 101 \(\times\) 51 - 51 \(\times\) 101 \(\times\) 45):51
=(1+5+7+...+2019) \(\times\) 0 : 51
= 0: 51
= 0
a) \(19\times\left(52+38\right)-9\times\left(52+38\right)=\)
\(=\) \(19\times90-9\times90\)
\(=90\times\left(19-9\right)\)
\(=90\times10\)
\(=900\)
b) \(\dfrac{2}{5}\div\dfrac{3}{7}\times\dfrac{3}{7}\div\dfrac{2}{5}+2019=\)
\(=\left(\dfrac{2}{5}\div\dfrac{2}{5}\right)\times\left(\dfrac{3}{7}\div\dfrac{3}{7}\right)+2019\)
\(=1\times1+2019\)
\(=2019\)
( 1 + 3 + 5 + 7 +...+ 2019) \(\times\) ( 4848 \(\times\) 75 - 7575 \(\times\) 48):51
=(1 + 3 + 5 + 7 +...+2019)\(\times\)( 48 \(\times\) 101 \(\times\) 75 - 75 \(\times\) 101 \(\times\)48):51
=(1+3+5+7+...+2019) \(\times\) ( 48 \(\times\) 101 \(\times\) 75 - 48 \(\times\) 101 \(\times\) 75):51
=(1+3+5+7...+2019)\(\times\)0 : 51
= 0: 51
= 0
a )
Ta có :
\(A=18\times19=\left(17+1\right)\times19=17\times19+19\)
\(B=17\times20=17\times\left(19+1\right)=17\times19+17\)
Do \(17\times19+19>17\times19+17\)
\(\Rightarrow A>B\)
Vậy \(A>B\)
b )
Ta có :
\(C=2019\times2019=\left(2018+1\right)\times2019=2018\times2019+2019\)
\(D=2018\times2020=2018\times\left(2019+1\right)=2018\times2019+2018\)
Do \(2018\times2019+2019>2018\times2019+2018\)
\(\Rightarrow C>D\)
Vậy \(C>D\)
mk ko viết đề ra nhé
a,2019 x (1 + 99)
=2019 x 100
=201900
c,2018 x (7 + 1 + 2)
=2018 x 10
=20180
mk chỉ bk phần a và c thôi hihi
a) \(2019+2019\times99=2019\times\left(1+99\right)=2019\times100=201900\)
b) \(\frac{7}{3}\times\frac{5}{9}+5\times\frac{3}{7}=\frac{35}{27}+\frac{15}{7}=\frac{650}{189}\)
c) \(2018\times7+2018+2018\times2=2018\times\left(7+1+2\right)=2018\times10=20180\)
d) \(\frac{225\times63}{9\times15}=\frac{15\times15\times9\times7}{9\times15}=15\times7=105\)