\(x\left(x+1\right)=2+4+6+...+2500\)

=>\(x\left(x+1\right)=2\cdot\left(1+2+3+...+1250\right)\)

=>\(x\left(x+1\right)=2\cdot\dfrac{1250\cdot1251}{2}=1250\cdot1251\)

=>x=1250

`x . (x + 1) = 2 + 4 +6 + ...+ 2500`

`=> x . (x + 1) = (2500 + 2) . [(2500 - 2) : 2 + 1] : 2`

`=> x . (x + 1) = 2502 . 1250 : 2`

`=> x . (x + 1) =1251. 1250`

`=> x . (x + 1) = (1250+1). 1250`

Do x là số nhiên, nên `x = 1250.`

Vậy `x = 1250 `

----------------------------

Phân tích thành nhân từ cũng được nhé:

`=> x . (x + 1) =1251. 1250`

`=> x^2 + x - 1251. 1250 = 0`

`=> x^2 + 1251x - 1250x - 1251. 1250 = 0`

`=> (x^2 + 1251x) - (1250x +  1251. 1250) = 0`

`=> x(x + 1251) - 1250 (x + 1251) = 0`

`=> (x-1250)(x+1251) = 0`

`=> x = 1250` hoặc `x = -1251`

Mà `x` là số tự nhiên nên `x = 1250`

+) 246a chia hết cho 2 

=> a ∈ {0; 2; 4; 6; 8} (1) 

+) 246a chia hết cho 3 

=> 2 + 4 + 6 + a = 12 + a chia hết cho 3 

12 ⋮ 3 => a ⋮ 3 => a ∈ {0; 3; 6; 9} (2) 

Từ (1) và (2) => a = 0 hoặc a = 6 

Để 246a là số lớn nhất thì a = 6 (vì 2460 < 2466) 

ABCD là hình chữ nhật 

=> CD = AB = 4 (cm) 

=> AD = BC = 3 (cm) 

=> BD = AC = 5 (cm)

1) 12 ⋮ x => x ∈ Ư(12) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 12; -12} 

Mà: x > 2 

=> x ∈ {3; 4; 6; 12} 

2) 24 ⋮ x => x ∈ Ư(24) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 8; -8; 12; -12; 24; -24} 

Mà: x > 4 

=> x ∈ {6; 8; 12; 24} 

3) 36 ⋮ x => x ∈ Ư(36) = {1; -1; 2; -2; 3; -3; 4; -4; 6; -6; 9; -9; 12; -12; 18; -18; 36; -36}

Mà: x ≥ 3 

=> x ∈ {3; 4; 6; 9; 12; 18; 36} 

4) 40 ⋮ x => x ∈ Ư(40) = {1; -1; 2; -2; 4; -4; 5; -5; 8; -8; 10; -10; 20; -20; 40; -40} 

Mà: x < 10 và x là số tự nhiên

=> x ∈ {1; 2; 4; 8}

Bài 1:

\(a.A=23,12+45,56+76,88+54,44\\ =\left(23,12+76,88\right)+\left(45,56+54,44\right)\\ =100+100\\ =200\\ b.201,5\cdot9+201,5\cdot2-201,5\\ =201,5\cdot\left(9+2-1\right)\\ =201,5\cdot10\\ =2015\\ c.C=\dfrac{1}{2}:0,5+\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+2014\\ =\dfrac{1}{2}\cdot2+\dfrac{1}{4}\cdot4+\dfrac{1}{8}\cdot8+2014\\ =1+1+1+2014\\ =2017\\ d.D=2\dfrac{2}{3}+\dfrac{5}{9}+\dfrac{4}{9}:\left(30\%-\dfrac{1}{10}\right)-\dfrac{2}{9}\\ =\dfrac{8}{3}+\left(\dfrac{5}{9}-\dfrac{2}{9}\right)+\dfrac{4}{9}:\left(\dfrac{3}{10}-\dfrac{1}{10}\right)\\ =\dfrac{8}{3}+\dfrac{1}{3}+\dfrac{4}{9}:\dfrac{1}{5}\\ =\dfrac{9}{3}+\dfrac{4}{45}\\ =3+\dfrac{4}{9}\cdot5=3+\dfrac{20}{9}=\dfrac{47}{9}\)

Bài 2: 

a: y+2=2017

=>y=2017-2=2015

b: \(3y-2\dfrac{2}{7}=3\dfrac{5}{7}\)

=>\(3y=3+\dfrac{5}{7}+2+\dfrac{2}{7}=6\)

=>\(y=\dfrac{6}{3}=2\)

c: \(1\dfrac{3}{4}-\dfrac{3}{4}y=75\%\)

=>\(\dfrac{7}{4}-\dfrac{3}{4}y=\dfrac{3}{4}\)

=>7-3y=3

=>3y=7-3=4

=>\(y=\dfrac{4}{3}\)

d: \(\dfrac{2}{3}+\dfrac{1}{3}y+3\dfrac{2}{3}y=\dfrac{8}{3}\)

=>\(4y=\dfrac{8}{3}-\dfrac{2}{3}=\dfrac{6}{3}=2\)

=>\(y=\dfrac{2}{4}=\dfrac{1}{2}\)

e: \(y-14=25\)

=>y=25+14=39

f: \(5y-25=35\)

=>5y=25+35=60

=>y=60/5=12

g: 9,34-y=1,28

=>y=9,34-1,28=8,06

h: y:1,2=2,4

=>\(y=2,4\cdot1,2=2,88\)

i: 2,4:y=0,2

=>y=2,4:0,2=12

k: (y+1)+(y+3)=24

=>y+1+y+3=24

=>2y=20

=>y=10

Nam đếm được 129 cái gì.

\(S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{200}}\)

\(\Rightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{200}}\right)\)

\(\Rightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{199}}\)

\(\Rightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{199}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{200}}\right)\)

\(\Rightarrow S=1-\dfrac{1}{2^{100}}< 1\)

\(\Rightarrow S< 1\)

Vậy \(S< 1\)