\(a)5 \times \dfrac{7}{5}=7\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(b)\dfrac{3}{4} \times \dfrac{7}{9}+\dfrac{1}{4} \times \dfrac{7}{9}\)

\(=\dfrac{7}{9} \times (\dfrac{3}{4}+\dfrac{1}{4})\)

\(=\dfrac{7}{9} \times \dfrac{4}{4}=\dfrac{7}{9}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(c)\dfrac{1}{7} \times \dfrac{5}{9}+\dfrac{5}{9} \times \dfrac{1}{7}+\dfrac{5}{9} \times \dfrac{3}{7}\)

\(=\dfrac{5}{9} \times (\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\)

\(=\dfrac{5}{9} \times \dfrac{5}{7}=\dfrac{25}{63}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(d)\dfrac{4 \times 11 \times 3}{4 \times 9 \times 121}\)

\(=\dfrac{4 \times 3 \times 11}{4 \times 3 \times 3 times 11 \times 11}\)

\(=\dfrac{1}{3 \times 11}=\dfrac{1}{33}\)

a, 5 x 7/5 = 7

b, 3/4 x 7/9 + 1/4 x 7/9

= 7/9 x ( 3/4 + 1/4 )

= 7/9 x 1

= 7/9

c, 1/7 x 5/9 + 5/9 x 1/7 + 5/9 x 3/7

= 1/7x ( 5/9+ 5/9) + 5/9 x 3/7

= 1/7 x 25/81 + 5/21

= 25/567 + 5/21

= 160/567

d, 4x11x3/4 x 9/121

= 44 x 3/4 x 9/121

= 33 x 9/121

= 27/11

57 + 725 + 605 - 5

= 782 + 600

= 1382

57 + 225 + 605 - 5

= 282 + 605 - 5

= 887 - 5

= 882

\(\dfrac{x-1}{2020}+\dfrac{x-2}{2021}=\dfrac{x-3}{2022}+\dfrac{x-4}{2023}\)

`=>`\(\dfrac{x-1}{2020}+1+\dfrac{x-2}{2021}+1=\dfrac{x-3}{2022}+1\dfrac{x-4}{2023}+1\)

`=>`\(\dfrac{x-1+2020}{2020}+\dfrac{x-2+2021}{2021}=\dfrac{x-3+2022}{2022}+\dfrac{x-4+2023}{2023}\)

`=>`\(\dfrac{x+2019}{2020}+\dfrac{x+2019}{2021}-\dfrac{x+2019}{2022}-\dfrac{x+2019}{2023}=0\)

`=>`\(\left(x+2019\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}\right)=0\)

`=>x+2019=0`

`=>x=-2019`

Vậy `x=-2019`

S = \(\dfrac{4n+5}{2n-3}\)  (n ϵ Z)

S = \(\dfrac{2\left(2n-3\right)+11}{\left(2n-3\right)}\)

S = 2 + \(\dfrac{11}{2n-3}\) 

S nguyên ⇔2n-3 ϵ Ư(11)  = {-11; -1; 1; 11}

⇔ n ϵ { -4; 1; 2; 7}

Ta có \(S=\dfrac{4n+5}{2n-3}=\dfrac{2\left(2n-3\right)+11}{2n-3}=2+\dfrac{11}{2n-3}\)

Để S lầ số nguyên =>\(\dfrac{11}{2n-3}\) nguyên

=> \(11⋮2n-3\) hay 2n-3 \(\in\) Ư(11)

  =>2n-3\(\in\) {1;11;-1;-11}

n\(\in\) {2;7;1;-4}

Lời giải:

a. $0,5.\frac{1}{2}+\frac{3}{4}=\frac{1}{4}+\frac{3}{4}=1$
b.

$\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{27.29}$

$=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{29-27}{27.29}$

$=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{29}$

$=1-\frac{1}{29}=\frac{28}{29}$

TL: 

X=\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2009}\)

Y=\(\dfrac{2008}{1}\)+\(\dfrac{2007}{2}\)+\(\dfrac{2006}{3}\)+....+\(\dfrac{2}{2007}\)+\(\dfrac{1}{2008}\)

Y=1+ (\(\dfrac{2007}{2}\)+1)+(\(\dfrac{2006}{3}\)+1)+....+(\(\dfrac{1}{2008}\)+1)

Y=\(\dfrac{2009}{2009}\)+\(\dfrac{2009}{2}\)+\(\dfrac{2009}{3}\)+.....+\(\dfrac{2009}{2007}\)+\(\dfrac{2009}{2008}\)

Y=2009(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2008}\)+\(\dfrac{1}{2009}\))

=> \(\dfrac{X}{Y}\)=\(\dfrac{1}{2009}\)

a, Gọi số chẵn là \(2k\) ta có: \(2k.3\) \(⋮\) \(6\)

\(2k.3=6k\); \(6\) \(⋮\) \(6\) \(\Rightarrow6k\) \(⋮\) \(6\) \(\Rightarrow2k.3\) \(⋮\) \(6\)

Vậy 3 số chẵn liên tiếp hay không liên tiếp thì vẫn chia hết cho 6 ( đpcm )

b, Gọi số lẻ là \(2k+1\) ta có: \(3\left(2k+1\right)\) \(⋮̸\)\(6\)

\(3\left(2k+1\right)=6k+3\); \(3\) \(⋮̸\)\(6\) \(\Rightarrow6k+3\) \(⋮̸\)\(6\Rightarrow3\left(2k+1\right)\) \(⋮̸\)\(6\)

Vậy 3 số lẻ liên tiếp hay không liên tiếp thì đều ko chia hết cho 6 ( đpcm )