Question

calculati:1) 22 inreg 1/3-(10 intreg 7/12+9intreg 11/20)
2)10intreg- (5intreg7/8+3intreg 5/12)
3)3intreg 11/18-(6intreg1/6-3inreg11/21)

va rog frumos !!!

Answer 1

1) 22 intreg 1/3-(10 intreg 7/12+9intreg 11/20)=
=67/3-127/12-191/20=
=(67*20-127*5-191*3)/60
=(1340-635-573)/60=
=132/60=11/5=2,2

2)10intregi- (5intreg7/8+3intreg 5/12)=
=10- 47/8-41/12=
=(24*10- 3*47/8-2*41)/24=
=(240- 141-82)/24=
=17/24=0,708(3)

3)3intreg 11/18-(6intreg1/6-3inreg11/21)=
=65/18-37/6-74/21=
=(7*65/18-21*37/6-6*74)/126=
=(455-777-444)/126=
=-766/126=
=-383/63=6,(079365)