`Ư(16) = {-16;-8;-4;-2;-1;1;2;4;8;16}`

`Ư(24) = {-24;-12;-8;-6;-4;-3;-2;-1;1;2;3;4;6;8;12;24}`

`=> ƯC(16;24) = {-8;-4;-2;-1;1;2;4;8}`

`Ư(30) = {-30;-15;-10;-6;-5;-3-2;-1;1;2;3;5;6;10;15;30}`

`Ư(45) = {-45;-15;-9;-5;-3;-1;1;3;5;9;15;45}`

`=> ƯC(30;45) = {-15;-5;-3;-1;1;3;5;15}`

Lời giải:

a. $140=2^2.5.7$

$168=2^3.3.7$

$\Rightarrow ƯCLN(140, 168) = 2^2.7=28$

b. 

$525=3.5^2.7$

$375=3.5^3$

$\Rightarrow ƯCLN(525, 375)=3.5^2=75$

Gọi số thứ nhất và số thứ hai lần lượt là: a,b
Trung bình cộng = a+b/2 = 285
=> a+b = 570
Ta có: 1/2 a = 1/3 b
=> a/2 = b/3
Áp dụng dãy tỉ số bằng nhau ta có: a/2 = b/3 = a+b/2+3 = 570/5 = 114
=> a = 2 x 114 = 228
=> b = 3 x 114 = 342
Vậy số thứ 1 = 228
       số thứ 2 = 342

Chỗ 570 = 285 x 2

là số:14,28 phải hum

Do `35 vdots x; 105 vdots x`

`=> x in UC{35;105)`

Mà `105 vdots 35`

`=> x in Ư(35) = {1;5;7;35}`

Mà `x > 5 -> x in {7;35}`

Vậy ...

`b) x vdots 10; x vdots 15`

`=> x in BC(10;15)`

Ta có: 

`10 = 2 . 5`

`15 = 3.5`

`=> BCN``N(10;15) = 2.3.5 = 30`

`=> x in B(30) = {0;30;60;90;120;...}`

Mà `x < 100 -> x in {0;30;60;90}`

`2/(1xx2xx3) + 2/(2xx3xx4) + ... + 2/(8xx9xx10) `

Ta có: 

`2/(1xx2xx3) = 1/(1xx2) - 1/(2xx3) `

`2/(2xx3xx4) = 1/(2xx3) - 1/(3xx4) `

....

`2/(8xx9xx10) = 1/(8xx9) - 1/(9xx10) `

Nên: `2/(1xx2xx3) + 2/(2xx3xx4) + ... + 2/(8xx9xx10) `

`= 1/(1xx2) - 1/(2xx3) + 1/(2xx3) - 1/(3xx4) + ... +  1/(8xx9) - 1/(9xx10) `

`= 1/2 - 1/90`

`= 22/45`

Vậy ....

\(\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\dfrac{2}{4\times5\times6}+...+\dfrac{2}{8\times9\times10}\)

\(=\dfrac{1}{1\times2}-\dfrac{1}{2\times3}+\dfrac{1}{2\times3}-\dfrac{1}{3\times4}+\dfrac{1}{3\times4}-\dfrac{1}{4\times5}+\dfrac{1}{4\times5}-\dfrac{1}{5\times6}+...+\dfrac{1}{8\times9}-\dfrac{1}{9\times10}\)

\(=\dfrac{1}{1\times2}-\dfrac{1}{9\times10}\)

\(=\dfrac{1}{2}-\dfrac{1}{90}\)

\(=\dfrac{22}{45}\)

a) Ta có: 

`84 = 2^2 . 3 . 7`

`108 = 2^2 . 3^3`

`=> UCLN(84;108) = 2^2 . 3 = 12`

`=> UC(84;108) = Ư(12) = {-12;-6;-4;-3;-2;-1;1;2;3;4;6;12}`

a) Tứ giác `ABCD` có `hat{A} + hat{B} + hat{C} + hat{D} = 360^o`

Do chúng lần lượt tỉ lệ với `2;3;6;7`

`=> hat{A}/2 = hat{B}/3 = hat{C}/6 = hat{D}/7`

Áp dụng t/chất dãy tỉ số bằng nhau:

`=>  hat{A}/2 = hat{B}/3 = hat{C}/6 = hat{D}/7 = (hat{A} + hat{B} + hat{C} + hat{D})/(2+3+6+7) = (360^o)/18 = 20^o`

`-> {(hat{A} = 20^o . 2 = 40^o),(hat{B} = 20^o  3 = 60^o),(hat{C} = 20^o . 6 = 120^o),(hat{D} = 20^o . 7 = 140^o):}`

Vậy ...

a; A =  5\(x^2\) + 10\(xy\) - 4\(x\) - 8y

   A = (5\(x^2\) + 10\(xy\)) - (4\(x\) - 8y) 

  A = 5\(x\).(\(x\) + 2\(y\)) - 4.(\(x+2y\))

 A =  (\(x+2y\)).(5\(x\) - 4)

B = 4\(x^2\) + 4\(x\) - y² + 1

B = (4\(x^2\) + 4\(x\) + 1) - y²

B = [(2\(x\))² + 2.2\(x\).1 + 1²] - y²

B = [2\(x\) + 1]² - y²

B = (2\(x+1\) - y)(2\(x+1\) - y)

\(a,\dfrac{5}{4}- \left(\dfrac{1}{2}\right)^2\)

\(=\dfrac{5}{4}-\dfrac{1}{4}\)

\(=1\)

\(b,\dfrac{2}{3}\cdot\dfrac{-3}{2}+\dfrac{-7}{2}\cdot\dfrac{2}{3}\)

\(=\dfrac{2}{3}\cdot\left(\dfrac{-3}{2}+\dfrac{-7}{2}\right)\)

\(=\dfrac{2}{3}\cdot-5\)

\(=-\dfrac{10}{3}\)

\(c,\left(\dfrac{1}{5}+\dfrac{4}{13}\right)+\left(\dfrac{-2}{5}+\dfrac{7}{13}\right)-\left(\dfrac{4}{5}-\dfrac{2}{13}\right)\)

\(=\dfrac{1}{5}+\dfrac{4}{13}-\dfrac{2}{5}+\dfrac{7}{13}-\dfrac{4}{5}+\dfrac{2}{13}\)

\(=\left(\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{4}{5}\right)+\left(\dfrac{4}{13}+\dfrac{7}{13}+\dfrac{2}{13}\right)\)

\(=-1+1\)

`=0`