已知方程x²+4x+m-1=0有两实数根x1,x2(1)求实属m的取值范围(2)当m=2时,求值①1/x1=1/x2 ②x1³+x2³
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![已知方程x²+4x+m-1=0有两实数根x1,x2(1)求实属m的取值范围(2)当m=2时,求值①1/x1=1/x2 ②x1³+x2³](/uploads/image/z/995909-5-9.jpg?t=%E5%B7%B2%E7%9F%A5%E6%96%B9%E7%A8%8Bx%26%23178%3B%2B4x%2Bm-1%3D0%E6%9C%89%E4%B8%A4%E5%AE%9E%E6%95%B0%E6%A0%B9x1%2Cx2%EF%BC%881%EF%BC%89%E6%B1%82%E5%AE%9E%E5%B1%9Em%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%EF%BC%882%EF%BC%89%E5%BD%93m%3D2%E6%97%B6%2C%E6%B1%82%E5%80%BC%E2%91%A01%2Fx1%3D1%2Fx2+%E2%91%A1x1%26%23179%3B%2Bx2%26%23179%3B)
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已知方程x²+4x+m-1=0有两实数根x1,x2(1)求实属m的取值范围(2)当m=2时,求值①1/x1=1/x2 ②x1³+x2³
已知方程x²+4x+m-1=0有两实数根x1,x2
(1)求实属m的取值范围(2)当m=2时,求值①1/x1=1/x2 ②x1³+x2³
已知方程x²+4x+m-1=0有两实数根x1,x2(1)求实属m的取值范围(2)当m=2时,求值①1/x1=1/x2 ②x1³+x2³
(1)
由方程有两个实数根,
∴△=b²-4ac=16-4(m-1)≥0,
∴ m≤5
(2)依据题意,根据韦达定理
x1+x2=-4
x1*x2=1
①是1/x1+1/x2吧
1/x1+1/x2
=(x1+x2)/(x1*x2)=-4
②x1³+x2³
=(x1+x2)*[x1²-x1x2+x2³]
=(x1+x2)*[(x1+x2)²-3x1x2]
=-4*(16-3)
=-52
1、方程有两实数根,可得:16-4(m-1)≥0解得:m≤52、当x=2时,原方程可化为:x²+4x+1=0根据根与系数关系可得:x1+x2=-4,x1x2=11/x1+1/x2=(x1+x2)/x1x2=-4x1³+x2³=(x1+x2)(x1²-x1x2+x2²) =(x1+x2)[(x1+x2)²-3x1x2] =-4(16-3) =-52