không biết giải

2001 

____

1991

1/1*2+1/2*3+1/3*4+1/4*5+...1/99+100

= 1/1x2 + 1/2x3 + 1/3x4 ...... +1/9x10

= 1-1/2+1/2-1/3+1/3-1/4+........+1/9-1/10

=1-1/10=9/10

đặt A=1/1 x 1/2 + 1/2 x 1/3 + 1/3 + 1/4 + .......... + 1/9 x 1/10

\(A=\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+...+\frac{1}{9}\cdot\frac{1}{10}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}\)

\(=\frac{9}{10}\)

đặt B=2/1 x 2 + 2/2 x 3 + 2/3 x4 + .............. + 2/98 x 99 + 2/99 x 100

\(B=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)

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

\(=2\left(1-\frac{1}{100}\right)\)

\(=2\times\frac{99}{100}\)

\(=\frac{99}{50}\)

\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{2010}{2009}=\frac{3.4.5...2010}{2.3.4....2009}=\frac{2010}{2}=1005\)

\(B=\frac{1.2.3......99}{1.2.3.4.....100}=\frac{1}{100}\)

G = 1.2 + 1 + 2.3 + 1 + 3.4 + 1 + ... + 99.100 + 1

= 1.2 + 2.3 + 3.4 + ... + 99.100 + 99

⇒ 3G = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3 + 99.3

= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)

= 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 + 297

= 99.100.101 + 297

= 1000197

⇒ G = 1000197 : 3

= 333399

vì tử của tất cả các số là 1-1 mà 1-1=0

suy ra:=0+0+0+...+0 (100 số 0)

Suy ra:=0

vậy (1-1/1+2).(1-1/1+2+3).....(1-1/1+2+3+...+99+100)=0

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: \(x=40\)

Ta có : 

\(P=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)

\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}....\frac{9898}{99.100}\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{98.101}{99.100}\)

\(P=\frac{1.2.3...98}{2.3.4....99}.\frac{4.5.6....101}{3.4.5...100}\)

\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)

Ủng hộ mk nha !!! ^_^

\(P=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)

\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)

\(P=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)

\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)

\(\frac{101}{297}\)\(nha\)\(bạn\)

M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)

M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

M = \(1-\dfrac{1}{100}\)

M = \(\dfrac{99}{100}\)