已知△ABC中,a²-16b²-c²+6ab+10bc=0(a,b,c是三角形的三边)求证:a+b=2b.
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已知△ABC中,a²-16b²-c²+6ab+10bc=0(a,b,c是三角形的三边)求证:a+b=2b.
已知△ABC中,a²-16b²-c²+6ab+10bc=0(a,b,c是三角形的三边)求证:a+b=2b.
已知△ABC中,a²-16b²-c²+6ab+10bc=0(a,b,c是三角形的三边)求证:a+b=2b.
∵a2-16b2-c2+6ab+10bc=a2+9b2+6ab-25b2-c2+10bc=(a+3b)2-(c-5b)2=0,
∴(a+3b+c-5b)(a+3b-c+5b)=0,
即(a+c-2b)(a-c+8b)=0,
∴a+c-2b=0或a-c+8b=0,
∴a+c=2b或a+8b=c,
∵a+b>c,
∴a+8b=c不符合题意,舍去,
∴a+c=2b.
a2+6ab+9b2-(25b2-10bc+c2)=0
(a+3b)2-(5b-c)2=0
因为a,b,c为三角形三边;
a+3b=5b-c;
a+c=2b
证明:a²-16b²-c²+6ab+10bc
=a²+6ab+9b²-(25b²-10bc+c²)
=(a+3b)²-(5b-c)²=0,
∴a+3b=5b-c或a+3b=-﹙5b-c﹚
∴a+c=2b或a+8b=c,
∵a+b>c,
∴a+8b=c不符合题意,舍去,
∴a+c=2b.