题目内容
已知,a-b=4,b+c=2,则a2+b2+c2-ab+bc+ca=( )
| A.56 | B.28 | C.24 | D.12 |
∵a-b=4①,b+c=2②,
∴①+②得:a+c=6,
∴a2+b2+c2-ab+bc+ca=
(2a2+2b2+2c2-2ab+2bc+2ca)
=
[(a2-2ab+b2)+(a2+2ac+c2)+(b2+2bc+c2)]
=
[(a-b)2+(a+c)2+(b+c)2]
=
×[42+62+22]
=
×56
=28.
故选B.
∴①+②得:a+c=6,
∴a2+b2+c2-ab+bc+ca=
| 1 |
| 2 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=
| 1 |
| 2 |
=28.
故选B.
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