∵a-b=2 b-c=3
∴a-c=5
a²+b²+c²-ab-bc-ac
=½(2a²+2b²+2c²-2ab-2bc-2ac)
=½[(a²+b²-2ab)+(a²+c²-2ac)+(b²+c²-2bc)]
=½[(a-b)²+(a-c)²+(b-c)²];
=½×(2²+5²+3²)
=½×(4+25+9)
=½×38
=19
∵a-b=2 b-c=3
∴a-c=5
a²+b²+c²-ab-bc-ac
=½(2a²+2b²+2c²-2ab-2bc-2ac)
=½[(a²+b²-2ab)+(a²+c²-2ac)+(b²+c²-2bc)]
=½[(a-b)²+(a-c)²+(b-c)²];
=½×(2²+5²+3²)
=½×(4+25+9)
=½×38
=19