va rog rezolvarea exercițiilor
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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!