Question

Am nevoie de ajutor la exercitiul 1, va rog.

Answer 1

1a)
=[3/4+2/3:(3/4-2/4)]·6/41
=[3/4+2/3:1/4]·6/41
=(3/4+2/3·4/1)·6/41
=(3/4+8/3)·6/41
=[(3·3+8·4)/12]·6/41
=41/12·6/41=6/12=simplif cu 6=1/2

b)
=[7/5·5/14+3/2:(9/4-3/2)]·2/7
=[1/2+3/2:(9/4-6/4)]·2/7
=[1/2+3/2:3/4]·2/7
=(1/2+3/2·4/3)·2/7
=(1/2+2)·2/7
=[(1+4)/2]·2/7
=5/2·2/7=5/7

c)
5intreg 2/3=(5·3+2)/3=17/3
1,(5)=(15-1)/9=14/9
15intreg 2/3=47/3
=[17/3+14/9-2]:47/3
=[(17·3+14-2·9)/9]·3/47
=[(51+14+18)/9]·3/47
=83/9·3/47=83/3·1/47=83/141

d)6intreg 2/3=20/3
2intreg1/4=9/4
0,(6)=6/9
0,(2)=2/9
4intreg5/6=29/6
7,25=725/100=simplif cu 25=29/4
={4+20/3-9/4·[6/9-2/9]}:29/6+29/4
={4+20/3-9/4:4/9}:29/6+29/4
={4+20/3+9/4·9/4}·6/29+29/4
={4+20/3+81/16}·6/29+29/4
={(4·48+20·16+81)/48}·6/29+29/4
={(192+320+81)/48}·6/29+29/4
=593/48·6/29+29/4
=593/8·1/29+29/4
=593/232+29/4
=(593+29·58)/232
=(593+1682)/232=2275/232