若x1,x2是方程x^2+2x-2009=0的两个根,试求下列各式的值(1)x1^2+x2^2 (2)1/x1+1/x2 (3)(x1-5)(x2-5) (4)|x1-x2| 注 :/是除 | |是绝对值 用伟达定理求
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![若x1,x2是方程x^2+2x-2009=0的两个根,试求下列各式的值(1)x1^2+x2^2 (2)1/x1+1/x2 (3)(x1-5)(x2-5) (4)|x1-x2| 注 :/是除 | |是绝对值 用伟达定理求](/uploads/image/z/5254286-14-6.jpg?t=%E8%8B%A5x1%2Cx2%E6%98%AF%E6%96%B9%E7%A8%8Bx%5E2%2B2x-2009%3D0%E7%9A%84%E4%B8%A4%E4%B8%AA%E6%A0%B9%2C%E8%AF%95%E6%B1%82%E4%B8%8B%E5%88%97%E5%90%84%E5%BC%8F%E7%9A%84%E5%80%BC%281%29x1%5E2%2Bx2%5E2+%282%291%2Fx1%2B1%2Fx2+%283%29%28x1-5%29%28x2-5%29+%284%EF%BC%89%7Cx1-x2%7C+%E6%B3%A8+%EF%BC%9A%2F%E6%98%AF%E9%99%A4+%7C+%7C%E6%98%AF%E7%BB%9D%E5%AF%B9%E5%80%BC+%E7%94%A8%E4%BC%9F%E8%BE%BE%E5%AE%9A%E7%90%86%E6%B1%82)
若x1,x2是方程x^2+2x-2009=0的两个根,试求下列各式的值(1)x1^2+x2^2 (2)1/x1+1/x2 (3)(x1-5)(x2-5) (4)|x1-x2| 注 :/是除 | |是绝对值 用伟达定理求
若x1,x2是方程x^2+2x-2009=0的两个根,试求下列各式的值
(1)x1^2+x2^2 (2)1/x1+1/x2 (3)(x1-5)(x2-5) (4)|x1-x2|
注 :/是除 | |是绝对值
用伟达定理求
若x1,x2是方程x^2+2x-2009=0的两个根,试求下列各式的值(1)x1^2+x2^2 (2)1/x1+1/x2 (3)(x1-5)(x2-5) (4)|x1-x2| 注 :/是除 | |是绝对值 用伟达定理求
根据一元二次方程根与系数关系得
X1+X2=-2,X1*X2=-2009
1)
X1^2+X2^2
=(X1+X2)^2-2*X1*X2
=4+4018
=4022
2)
1/X1+1/X2
=(X1+X2)/(X1*X2)
=2/2009
3)
(X1-5)(X2-5)
=X1*X2-5(X1+X2)+25
=-2009-5(-2)+25
=1974
4)
|X1-X2|
=√(X1-X2)^2
=√[(X1+X2)^2-4X1*X2]
=√(4+4*2009)
=√8040
=2√2010