Question

va rog rezolvarea exercițiilor ​

Answer 1

Răspuns:

Explicație pas cu pas:

11.

a) (x+1)+(x+2)+...+(x+25)=100  ⇒25·x+1+2+...+25=100  ⇒25·x+25·26/2=100  ⇒25·x+25·13=100  ⇒25·(x+13)=100 |:25  ⇒x+13=4  ⇒x=-9

b) (x+1)+(x+2)+...+(x+23)=437  ⇒23·x+1+2+...+23=437  ⇒23·x+23·24/2=437  ⇒23·x+23·12=437  ⇒23·(x+12)=437 |:23 ⇒x+12=19  ⇒x=7

c) (x+1)+(x+2)+...+(x+33)=693 ⇒33·x+1+2+...+33=693  ⇒33·x+33·34/2=693  ⇒33·x+33·17=693  ⇒33·(x+17)=693 }:33  ⇒x+17=21  ⇒x=4

d) Acum ai vazut cum se face acest tip de exercitiu, subpunctul d) il las

sa-l faci tu. Am folosit formula sumei lui Gauss: 1+2+3+...+n=n·(n+1)/2.

12.

a) 1/5·{1/5·[1/5·(1·x/5 + 2)+2]+2}=1  ⇒{1/5·[1/5·(1·x/5 + 2)+2]+2}=5  ⇒

1/5·[1/5·(1·x/5 + 2)+2]+2=5  ⇒1/5·[1/5·(1·x/5 + 2)+2]=5-2  ⇒1/5·[1/5·(1·x/5 + 2)+2]=3  ⇒[1/5·(1·x/5 + 2)+2]=3·5  ⇒1/5·(1·x/5 + 2)+2=15  

⇒1/5·(1·x/5 + 2)=15-2   ⇒1/5·(1·x/5 + 2)=13 ⇒(1·x/5 + 2)=13·5 ⇒(1·x/5 + 2)=65 ⇒1·x/5=65-2  ⇒1·x/5=63  ⇒x=63·5  ⇒x=315

b)  1/5·{1/4·[1/3·(1·x/2 - 1)-2]-3}=1  ⇒1/4·[1/3·(1·x/2 - 1)-2]-3=5  ⇒

1/4·[1/3·(1·x/2 - 1)-2]=8  ⇒[1/3·(1·x/2 - 1)-2]=8·4  ⇒1/3·(1·x/2 - 1)-2=32  ⇒1/3·(1·x/2 - 1)=32+2  ⇒1/3·(1·x/2 - 1)=34  ⇒(1·x/2 - 1)=34·3  ⇒1·x/2 - 1=102  ⇒

1·x/2=102+1  ⇒1·x/2=103  ⇒x=103·2  ⇒x=206

c) 1/4·{1/4·[1/4·(1·x/4 + 3)+3]+3}=1  ⇒{1/4·[1/4·(1·x/4 + 3)+3]+3}=1·4  ⇒1/4·[1/4·(1·x/4 + 3)+3]+3=4  ⇒1/4·[1/4·(1·x/4 + 3)+3]=4-3  ⇒

1/4·[1/4·(1·x/4 + 3)+3]=1  ⇒[1/4·(1·x/4 + 3)+3]=1·4  ⇒1/4·(1·x/4 + 3)+3=4  ⇒

1/4·(1·x/4 + 3)=4-3  ⇒1/4·(1·x/4 + 3)=1  ⇒(1·x/4 + 3)=1·4  ⇒1·x/4 + 3=4  ⇒

1·x/4=4-3  ⇒x/4=1  ⇒x=4

d) Ai vazut cum se face acest tip de exercitiu, subpunctul d) il las

sa-l faci tu.

Succes!