can you help me thank you very much .已知▲ABC的三条边为a、b、c、,方程x²+2(a+c-b)x+2ac-b²=0有两个相 等 的实数根.（1）求证：▲ABC是等边三角形；（2）若方程2x²+2k(a+c)x+2ack²+5k-6=0有两个不

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can you help me thank you very much .已知▲ABC的三条边为a、b、c、,方程x²+2(a+c-b)x+2ac-b²=0有两个相 等 的实数根.（1）求证：▲ABC是等边三角形；（2）若方程2x²+2k(a+c)x+2ack²+5k-6=0有两个不

can you help me thank you very much .已知▲ABC的三条边为a、b、c、,方程x²+2(a+c-b)x+2ac-b²=0有两个相 等 的实数根.（1）求证：▲ABC是等边三角形；（2）若方程2x²+2k(a+c)x+2ack²+5k-6=0有两个不
can you help me thank you very much .
已知▲ABC的三条边为a、b、c、,方程x²+2(a+c-b)x+2ac-b²=0有两个相 等 的实数根.（1）求证：▲ABC是等边三角形；（2）若方程2x²+2k(a+c)x+2ack²+5k-6=0有两个不相等的实数根,求k的取值范围.

can you help me thank you very much .已知▲ABC的三条边为a、b、c、,方程x²+2(a+c-b)x+2ac-b²=0有两个相 等 的实数根.（1）求证：▲ABC是等边三角形；（2）若方程2x²+2k(a+c)x+2ack²+5k-6=0有两个不
1、(2(a+c-b))^2-4(2ac-b^2)=0
化简得到(a-b)^2+(b-c)^2=0,由三条边都为正实数,因而a=b=c
2、[2k(a+c)]^2-8(2ack^2+5k-6)>0
用上1的结论,可得k

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