已知;x1,x2是一元二次方程x²-4x+1=0的两个实数根,求(x1+x2)²÷﹙1/x1+1/x2)的值
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已知;x1,x2是一元二次方程x²-4x+1=0的两个实数根,求(x1+x2)²÷﹙1/x1+1/x2)的值
已知;x1,x2是一元二次方程x²-4x+1=0的两个实数根,求(x1+x2)²÷﹙1/x1+1/x2)的值
已知;x1,x2是一元二次方程x²-4x+1=0的两个实数根,求(x1+x2)²÷﹙1/x1+1/x2)的值
解
x1,x2是方程根
∴有韦达定理得:
x1+x2=4
x1x2=1
∴(x1+x2)²÷(1/x1+1/x2)
=16÷[(x1+x2)/(x1x2]
=16÷(4/1)
=16÷4
=4
x1+x2=4
x1x2=1
所以原式=(x1+x2)²/[(x1+x2)/x1x2]
=x1x2(x1+x2)
=4