dla jakich x liczby 4x-5, x2-4, 2x+17 sa trzema kolejnymi liczbami ciagu arytmetycznego
a1= 4x-5
a2=2x-4
a3=2x+17
r= a3-a2=2x+17-(2x-4)=2x+17-2x+4=21
a3= a1+2r= 4x-5 + 2*21=4x-5+42=4x+37
4x+37=2x+17
4x-2x=17-37
2x=-20 |:2
x= -10
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a1= 4x-5
a2=2x-4
a3=2x+17
r= a3-a2=2x+17-(2x-4)=2x+17-2x+4=21
a3= a1+2r= 4x-5 + 2*21=4x-5+42=4x+37
4x+37=2x+17
4x-2x=17-37
2x=-20 |:2
x= -10