x²+(根号2+根号3)x+根号6=0
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![x²+(根号2+根号3)x+根号6=0](/uploads/image/z/1825537-49-7.jpg?t=x%26%23178%3B%2B%EF%BC%88%E6%A0%B9%E5%8F%B72%2B%E6%A0%B9%E5%8F%B73%EF%BC%89x%2B%E6%A0%B9%E5%8F%B76%3D0)
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x²+(根号2+根号3)x+根号6=0
x²+(根号2+根号3)x+根号6=0
x²+(根号2+根号3)x+根号6=0
b²-4ac=(根号3+根号2)²-4根号6=5+2根号6-4根号6=5-2根号6=(根号3-根号2)²
所以
x1=[(根号3+根号2)+(根号3-根号2)]/2
=2根号3/2
=根号3
x2=[(根号3+根号2)-(根号3-根号2)]/2
=2根号2/2
=根号2