一元二次方程ax^2+bx+c=0(a不等于0)的两个实根为x1,x2. 求（1）(x1-x2)的绝对值 （2）x1^3+x2^3

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解：1）原式=√枯友纳[(x1-x2)²]
=√[(x1+x2)²-4x1x2]
因为x1+x2=-b/a,x1x2=c/a
所以原式告冲=√(b²/a²-4c/a)
2）原式=(x1+x2)³-3x1x2(x1+x2)
因为没没x1+x2=-b/a,x1x2=c/a
所以原式=-b³/a³+3bc/a²

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ax^2+bx+c=0(a不等于0)

x1x2=c/a
x1+x2=-b/a

（1）(x1-x2)的绝对值

=√(x1-x2)²=√[(x1+x2)²-4x1x2]=[√(b²-4ac)]/|a|

（2）x1^3+x2^3

=(x1+x2)³尘裂歼-3x1x2(x1+x2)
=-b³/a³源粗-3c/a* (-b/a)
=-b³/a³派冲+3bc/a²

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用和迹韦达定理
x1+x2=-b/a
x1*x2=c/a
所以
(1)
(x1-x2)²= （x1+x2）²-4x1x2=(b²-4ac)/a²
所以租察（x1-x2)的绝对值=√唤型并(b²-4ac)/|a|
(2)
x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)
=(x1+x2)[(x1+x2)^2-3x1x2]
=(-b³+abc)/a³

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(x1-x2)²
=(x1+x2)²-4x1x2
所以培瞎逗|x1-x2|=√(b²/a²-4c/a)

x1²+x2²=(x1+x2)²-2x1x2
所以原配卖式=(x1+x2)(x1²神知-x1x2+x2²)
=-b/a*(b²/a²-3c/a)
=-b(b²-3ac)/a³

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(1)
|x1-x2|=根号困升下|x1-x2|^2=根号下(x1+x2)^2-4x1x2
x1+x2=-b x1x2=c
所以|x1-x2|=根号下| (x1+x2)^2-4x1x2|=根汪神老号下|b^2/a^2-4c/a|=根瞎碧号下|b^2-4ac|/|a|
(2)
x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)
=(x1+x2)(x1^2-x1x2+x2^2+2x1x2-2x1x2)
=(x1+x2)((x1+x2)^2-3x1x2)
=-b/a((-b/a^2-3c/a)
=(abc-b^3)/a^3

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