Question

a²+b²+c²=14 若a=b+c 求代数式ab-bc+ac的值

Answer 1

a=b+c的两边平方，
得a²=b²+c²+2bc (1)
又因为a²+b²+c²=14 (2)
上面二式(1),(2)相加得：2a²=14+2bc
得a²-bc=7.
故ab-bc+ac=a(b+c)-bc=a²-bc=7。

Answer 2

a=3,b=2,c=1
ab-bc+ac=7

Answer 3

(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc=0
ab-bc+ac=(a^2+b^2+c^2)/2=7

Answer 4

a^+b^+c^
=(b+c)^+b^+c^
=b^+c^+2bc+b^+c^
=2(b^+c^+bc)
=14

=>b^+c^+bc=7

ab-bc+ac
=(b+c)b-bc+(b+c)c
=(b+c-c)b+bc+c^
=b^+bc+c^
=7

Answer 5

解:∵a=b+c→a-b-c=0
∴（a-b-c）²=a²+b²+c²-2ab+2bc-2ac=0
∴2(ab-bc+ac)=a²+b²+c²=14
∴(ab-bc+ac)=7

Answer 6

ab-bc+ac
=a(b+c)-bc
=a*a-bc a2+b2+c2=14
a2+(b+c)2-2bc=14
bc=a2+（b+c）2-14/2
代入：a*a-bc
=a2-[a2+(b+c)2-14/2]
=a2-[a2+a2-14/2]
=a2-a2+7
=7