x+(x+1)+(x+2)+...(x+100)=8080

(x+x+x+..+x)+(0+1+2+...+100)=8080

101x+5050=8080

101x=8080-5050

101x=3030

x=3030:101

x=30

x + ( x + 1 ) + ( x + 2 ) + ( x + 3 ) +.....+ ( x + 100 ) = 8080

=( x + x + x + x +....+ x ) + ( 1 + 2 + 3 + 4 + .... + 100) = 8080

=(x + x + x + ... + x ) + 5050 = 8080

 101x = 8080 - 5050

101x = 3030

x = 3030 : 101

x = 30

=(x + x + x + ... + x ) + 5050 = 8080

 101x = 8080 - 5050

101x = 3030

x = 3030 : 101

x = 30

= 2020 x 0,56 + 0,34 x 3 x 2020 + 2020 x 4 x 0,85 + 2020 x 5,02

= 2020 x 0,56 + 1,02 x 2020 + 2020 x 3,4 + 2020 x 5,02

= 2020 x ( 0,56 + 1,02 + 3,4 + 5,02 )

= 2020 x ( 1,58 + 3,4 + 5,02 )

= 2020 x ( 4,98 + 5,02 )

= 2020 x 10

= 20202

         ------------------Hết------------------       

Tính theo tính nhanh_________

tui nhầm thay 2 ở cuối thành 0

 2020 x 0,56 + 0,36 x 6060 + 8080 x 0,85 + 2020 x 5,02 

=1131,2+2181,6+6868+10140,4 

=20321,2

   2020 x 0,56 +0,36x6060+8080x0,85+2020x5,02

=2020 x 0,56 +0,36x3x2020+4x2020x0,85+2020x5,02

=2020 x 0,56 +1,08x2020+3,4x2020+2020x5,02

=2020x(0,56+1,08+3,4+5,02)

= 2020x10,06

=20321.2

a: S=1(1+1)+2(1+2)+...+100(1+100)

=1+2+...+100+1^2+2^2+...+100^2

\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)

\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)

=343450

b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)

=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)

=>4*A=100*101*102*103

=>A=25*101*102*103

giúp mik nhé cám ơn

16

=28

a)x+(x+1)+(x+2)+(x+3)+...+(x+99)+(x+100)=5555

=> 101x +5050 = 5555

=> 101x = 505

=> x = 505 : 101 = 5

Vậy, x = 5

b)1+2+3+4+...+x=820

=> ( x+1) x :2 = 820

=> (x+1)x = 1640

Mà 1640 = 40 . 41

=> x = 40 ( vì {x+1} - x = 1)

Vậy, x = 40

c) 3^x+1 = 9.27=243

=> 3^x+1 = 3⁵

=>x + 1 = 5

=> x = 4

Vậy, x=4

d) x+2x+3x+...+99x+100x=15150

=> [( 100 + 1) x 100 :2 ] x = 15150

=> 5050x = 15150

=> x = 15150:5050 = 3

Vậy, x =3

e)(x+1)+(x+2)+(x+3)+...+(x+100)=205550

=> 100x + 5050 = 205550

=> 100x =  205550 - 5050= 200500

=> x =  200500 : 100 = 2005

Vậy, x = 2005

f)3^x+3^x+1+3^x+2=351

=> 3^x + 3^x . 3 + 3^x x 9 = 351

=> 3^x ( 1+3+9) = 351

=> 3^x  . 13 = 351

=> 3^x  = 351 :13=27 mà 27 = 3³

=> x=3

Vậy, x=3

mình đg cần gấp á

a) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+100\right)=5555\)

\(\Rightarrow x+x+1+x+2+x+3+...+x+100=5555\)

\(\Rightarrow101\cdot x+5050=5555\)

\(\Rightarrow101\cdot x=5555-5050\)

\(\Rightarrow101\cdot x=505\)

\(\Rightarrow x=505:101\)

\(\Rightarrow x=5\)

b) \(1+2+3+4+...+x=820\)

\(\Rightarrow\left(x+1\right)\cdot\left[\left(x-1\right):1+1\right]:2=820\)

\(\Rightarrow\left(x+1\right)\cdot\left(x+1-1\right):2=820\)

\(\Rightarrow\left(x+1\right)\cdot x:2=820\)

\(\Rightarrow x\cdot\left(x+1\right)=820\cdot2\)

\(\Rightarrow x\cdot\left(x+1\right)=1640\)

Ta thấy: \(40\cdot41=1640\)

Vậy: \(x=40\)

      A=100/1 x 2 + 100/2 x 3 + 100/3 x 4 +...+100/99 x 100

A/100=1/1 x 2 + 1/2 x 3 + 1/3 x 4 +...+1/99 x 100

A/100=2-1/1x2 + 3-2/2x3 + ... + 100-99/99x100

A/100=1-1/2 + 1/2-1/3+...+1/99-1/100

A/100=1-1/100

A/100=99/100

A=99/100x100=99

Vậy A=99.

Ta có:

\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)

\(\Rightarrow100.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(\Rightarrow100.\left(\frac{1}{1}-\frac{1}{100}\right)\Leftrightarrow100.\frac{99}{100}=99\)

\(\text{Ta có :}\)

\(\frac{100}{1.2}+\frac{100}{2.3}+\frac{100}{3.4}+...+\frac{100}{99.100}\)

\(=100.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(100.\left(\frac{1}{1}-\frac{1}{100}\right)=100.\frac{99}{100}=99\)