1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x + 1) = 99/100
1- 1/2 +1/2-1/3+1/3-1/4+...+ 1/x - 1/ x+ 1 = 99/100
1 - 1/ x+1 = 99/ 100
=> (100 - 1)/ x+1 = 99 / 100
=> x+1 = 100 => x=99

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\)

a, => 3x-17 = 0 hoặc 3x-17 = 1

=> x=17/3 hoặc x=6

b, => x+1+x+2+....+x+100=205550

=>100x + (1+2+...+100)=205550

=> 100x + 5050 = 205550

=> 100x = 205550 - 5050 = 200500

=>x= 2005

c,=>x+x+1+....+x+2010=2029099

=>2011x+(1+2+....+2010)=2029099

=>2011x+2021055=2029099

=>2011x = 2029099-2021055 = 8044

=>x=4

Có : 3Q = 3+3^2+....+3^101

2Q=3Q-Q= (3+3^2+....+3^101)-(1+3+3^2+...+3^100) = 3^101-1

=>Q = (3^101-1)/2

cho \(M=1+3+3^2+...+3^{99}+3^{100}\)

=>\(M=1+\left(3+3^2+3^3\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(=>M=1+3\left(1+3+3^2\right)+...+3^{98}\left(1+3+3^2\right)\)

\(=>M=1+13\left(3+...+3^{98}\right)\)

Mà \(13\left(3+3^{98}\right)⋮13\)

=> M chia cho 13 dư 1

+) \(M=1+3+3^2+...+3^{99}+3^{100}\)

\(\Leftrightarrow M=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{98}+3^{99}+3^{100}\right)\)

\(\Leftrightarrow M=\left(1+3+9\right)+3^3\left(1+3+9\right)+....+3^{98}\left(1+3+9\right)\)

\(\Leftrightarrow M=13+3^3\cdot14+....+3^{98}\cdot14\)

\(\Leftrightarrow M=13\left(1+3^3+....+3^{98}\right)\)

=> M chia 13 dư 0

vì |x+1| lớn hơn hoặc bằng 0, |x+2| lớn hơn hoặc 0 suy ra |x+1|+|x+2| lớn hơn hoặc bằng 0 nên |x+1|+|x+2|=0

suy ra |x+1|=0

           |x+2|=0

suy ra x+1=0

           x+2=0

xong tìm x đi nhé chúc bạn học giỏi 

Vì |x+1| và |x+2| đều >=0 => |x+1|+|x+2| >=0

Mà |x+1|+|x+2|=0 => |x+1|+|x+2| = 0 <=> x+1 = 0 và x+2 = 0

<=> x=-1 và x=-2 ( vô lý )

=> ko tồn tại x thỏa mãn bài toán trên

k mk nha

A=9954

????????????????????

x=(1+2+3-4-5-6)+...+(97+98+99-100-101-102)

x=-9+...+-9

x=-9.17

x=-153

$A=\dfrac12-\dfrac{2}{2^2}+\dfrac{3}{2^3}-\dfrac{4}{2^4}+\cdots+\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}$

Nhóm từng 2 số:

$A=\left(\dfrac12-\dfrac{2}{2^2}\right)+\left(\dfrac3{2^3}-\dfrac4{2^4}\right)+\cdots+\left(\dfrac{99}{2^{99}}-\dfrac{100}{2^{100}}\right)$

$=0+\dfrac18+\dfrac{2}{32}+\dfrac3{128}+\cdots+\dfrac{49}{2^{99}}$

$=\sum_{k=1}^{50}\dfrac{k-1}{2^{2k-1}}$

Ta có: $\dfrac{k-1}{2^{2k-1}}=\dfrac{2(k-1)}{4^k}$

Mà: $\sum_{k=1}^{\infty}\dfrac{k-1}{4^k}=\dfrac{1}{9}$

Nên: $A<2\cdot\dfrac19$ $=\dfrac29$

Vậy: $A<\dfrac29$

$4=1\cdot4,\quad28=4\cdot7,\quad70=7\cdot10,\ldots$

tức là: $E=\dfrac3{1\cdot4}+\dfrac3{4\cdot7}+\dfrac3{7\cdot10}+\cdots+\dfrac3{n(n+3)}$

Với $n=1,4,7,\ldots$.

Ta có: $\dfrac3{n(n+3)}=\dfrac1n-\dfrac1{n+3}$

Do đó: $E=\left(1-\dfrac14\right)+\left(\dfrac14-\dfrac17\right)+\left(\dfrac17-\dfrac1{10}\right)+\cdots+\left(\dfrac1n-\dfrac1{n+3}\right)$

$=1-\dfrac1{n+3}$

Vì: $\dfrac1{n+3}>0$ nên: $1-\dfrac1{n+3}<1$

Chieu nay nhek

Đặt B=2+2^2+2^3...+2^100

2B=2^2+2^3+2^4+.....+2^101

2B-B=2^2+2^3+...+2^101-2-2^2-...-2^100

B=2^101-2

Ta có:2^n-1-2-2^2-2^3-...-2^100=1

<=>2^n-1-(2+2^2+...+2^100)=1

<=>2^n-1-B=2^n-1-(2^101-2)=1

<=>2^n-(2^101-2)=2

<=>2^n=2+2^101-2=2^101

<=>n=101

Vậy n=101