设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要条
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设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要条
设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要
设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要条件是角A等于90°
设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要设a,b,c为△ABC的三边,求证:关于x的方程x²+2ax+b²=0与x²-2cx-b²=0有公共根的充要条
充分性
因为A等于90°
所以b2+c2=a2代入x²+2ax+b²=0与x²-2cx-b²=0中得x²+2ax+ a2- c2=0(x+a)2= c2
得到
x²-2cx-b²=0得到x²-2cx- a2+ c2=0得到(x-c)2= a2
得出公共根.
必要性
先解方程x²+2ax+b²=0与x²-2cx-b²=0得到x=(-b2)/(a+c),代入第一个方程求出勾股定理的式子
必要性:第一个方程减去第二个方程解得x=(-b^2)/(a+c),代入第二个方程,整理化简立得b^2+c^2=a^2。
充分性:显然。
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