Question

va rog repede........​

Answer 1

$(\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{99*100}):[(\frac{3}{5})^4*(\frac{3}{5})^3:(\frac{3}{5})^5]=\\=(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}):[(\frac{3}{5})^7:(\frac{3}{5})^5]=\\=(1-\frac{1}{100}):[(\frac{3}{5})^2]=$
$=\frac{99}{100}:\frac{9}{25}=\frac{99}{100}*\frac{25}{9}=\frac{11}{4}$

Answer 2

Răspuns:

Explicație pas cu pas:

[1/(1·2)+1/(2·3)+ ... + 1/(99·100)]:[(3/5)⁴· (3/5)³:(3/5)⁵]

1/(1·2)=1/1-1/2=(2-1)/1·2

1/(2·3)=1/2-1/3=(3-2)/(2·3)

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1/(99·100)=1/99-1/100=(100-99)/(99·100)

Acum se inlocuiesc fractiile cu aceste diferente de fractii scrise cu bold si se poate vedea ca toate fractiile se reduc, cu exceptioa primei fractii si ultimei fractii. ⇒[1/(1·2)+1/(2·3)+ ... + 1/(99·100)]:[(3/5)⁴· (3/5)³:(3/5)⁵]=

(1/1-1/2+1/2-1/3+ .... + 1/99-1/100):[(3/5)⁷:(3/5)⁵] =

(1/1-1/100):(3/5)²= (100-1)/100:9/25=99/100·25/9=

(99·25)/(100·9)= (11·25)/100= 11/4