1B

2B

3B

4A

5A

1B

2C

3A

4B

`19dm^3=19 : 1000= 0,019 m^3`

`5m^3 19dm^3=5+0,019=5,019m^3`

`5m^3 19dm^3 =5+19:1000=5,019m^3`

Sai nha bạn

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#Bonnie @Ngân Hà

989999901 nha

9999999 x 999 = 9989999001 

 bằng 989999901

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MÌNH CHẬM LẮM KO GIÚP ĐC

Có ai giải được không?

mình giải đc

11: \(27+12+\left(-17\right)+\left(-12\right)\)

\(=\left(27-17\right)+\left(12-12\right)\)

=10+0

=10

12: \(-564-\left(324-564-224\right)\)

\(=-564-\left(-564+100\right)\)

\(=-564+564-100=-100\)

13: \(2^3\cdot3^2-7^2-2022^0\)

\(=8\cdot9-49-1\)

=72-50

=22

14: \(236\cdot145+236\cdot856-236\)

\(=236\cdot145+236\cdot856-236\cdot1\)

\(=236\left(145+856-1\right)\)
\(=236\cdot1000=236000\)

15: \(-\left(-357\right)+27+\left(-357\right)+\left(-32\right)\)

\(=357+27-357-32\)

=27-32

=-5

16: \(275\cdot48+22\cdot275+175\left(-70\right)\)

\(=275\left(48+22\right)-175\cdot70\)

\(=275\cdot70-175\cdot70\)

\(=70\left(275-175\right)=70\cdot100=7000\)

17: \(\dfrac{\left(23\cdot36-17\cdot36\right)}{36}\)

\(=\dfrac{36\left(23-17\right)}{36}\)

=23-17

=6

18: \(3\cdot5^3-27:3^2+5^3\cdot4-18:3^2\)

\(=3\cdot5^3+4\cdot5^3-27:9-18:9\)

\(=5^3\left(3+4\right)-3-2\)

\(=125\cdot7-5=875-5=870\)

19: \(-87+\left(-12\right)-\left(-487\right)+512\)

\(=-87-12+487+512\)

\(=\left(487-87\right)+\left(512-12\right)\)

=400+500

=900

20: \(25\cdot57+75+33\cdot25+75\cdot89\)

\(=25\cdot57+25\cdot33+75+75\cdot89\)

\(=25\left(57+33\right)+75\left(1+89\right)\)

\(=25\cdot100+100\cdot75\)

\(=100\left(25+75\right)=100\cdot100=10000\)

a) \(\dfrac{2}{5}\) với \(\dfrac{3}{4}\)

Ta có : \(2\times4=8\) 

           \(5\times3=15\)

Mà \(8< 15\)

Vậy \(\dfrac{2}{5}< \dfrac{3}{4}\)

b) \(\dfrac{4}{9}\) và \(\dfrac{3}{10}\)

\(4\times10=40\)

 \(3\times9=27\)

Mà \(40>27\)

Vậy \(\dfrac{4}{9}>\dfrac{3}{10}\)

a) <

b) > 

a: Độ dài quãng đường AB là 2*(30+50)=160km

b: Chỗ gặp cách A:

2*30=60km