Đặt \(A=\overline{456xy}\)

Để A chia hết cho 9 thì \(x+y+4+5+6⋮9\)

=>\(x+y+15⋮9\)

=>x+y chia 9 dư 3
Để A chia hết cho 4 thì \(\overline{xy}⋮4\)

=>(x;y)\(\in\){(0;4);(0;8);(1;2);(1;6);(2;0);(2;4);(2;8);(3;2);(3;6);(4;0);(4;4);(4;8);(5;2);(5;6);(6;0);(6;4);(6;8);(7;2);(7;6);(8;0);(8;4);(8;8);(9;2);(9;6);(0;0)}

mà x+y chia 9 dư 3

nên (x;y)\(\in\){(4;8);(8;4)}

=>Có 2 cặp

Có 3 cặp, là 45648; 45684; và 45612

\(\dfrac{3}{4}+\dfrac{2}{3}=\dfrac{9+8}{12}=\dfrac{17}{12}\)

còn mấy câu kia nữa bạn

\(\overline{abc}=\overline{ac}\times6+15\)

=>\(100a+10b+c=6\left(10a+c\right)+15\)

=>\(40a+10b-5c=15\)

=>8a+2b-c=15

mà a,b,c là các số tự nhiên có 1 chữ số và a<>0

nên a=1; b=4; c=1

c 

180m/giây=0,18km/1/60 giờ=10,8km/h

0,25x6,117x40-1,117x1,25x8

=10x6,117-1,117x10

=10(6,117-1,117)

=10x5=50

4,16

4,61

1,64

1,46

6,14

6,41

6,25x2+6,25x25%+6,25+6,25x75/100+6,25x6

=6,25x2+6,25x0,25+6,25+6,25x0,75+6,25x6

=6,25(2+0,25+1+0,75+6)

=6,25x10=62,5

3,7x5,8+4,1x3,7+0,1x3,7

=3,7x(5,8+4,1+0,1)

=3,7x10=37

\(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+3+...+20}\)

\(=\dfrac{1}{2\cdot\dfrac{3}{2}}+\dfrac{1}{3\cdot\dfrac{4}{2}}+...+\dfrac{1}{20\cdot\dfrac{21}{2}}\)

\(=\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{20\cdot21}\)

\(=2\left(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{20\cdot21}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{21}\right)=2\cdot\dfrac{19}{21}=\dfrac{38}{21}\)