问题
解答题
附加题:已知a+b=5,b+c=2,求多项式a2+b2+c2+ab+bc-ac的值.
答案
∵a+b=5①,b+c=2②
∴①-②得a-c=3
∴a2+b2+c2+ab+bc-ac
=
(2a2+2b2+2c2+2ab+2bc-2ac)1 2
=
[(a+b)2+(b+c)2+(a-c)2]1 2
=
(25+4+9)1 2
=19.
附加题:已知a+b=5,b+c=2,求多项式a2+b2+c2+ab+bc-ac的值.
∵a+b=5①,b+c=2②
∴①-②得a-c=3
∴a2+b2+c2+ab+bc-ac
=
(2a2+2b2+2c2+2ab+2bc-2ac)1 2
=
[(a+b)2+(b+c)2+(a-c)2]1 2
=
(25+4+9)1 2
=19.