(3x+1)²-(4x+13)²=0
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![(3x+1)²-(4x+13)²=0](/uploads/image/z/1333272-48-2.jpg?t=%EF%BC%883x%2B1%EF%BC%89%26%23178%3B-%EF%BC%884x%2B13%EF%BC%89%26%23178%3B%3D0)
(3x+1)²-(4x+13)²=0
(3x+1)²-(4x+13)²=0
(3x+1)²-(4x+13)²=0
利用平方差公式
(3x+1)²-(4x+13)²
=(3x+1-4x-13)(3x+1+4x+13)
=(-x-12)(7x+14)=0
所以x=-12或-2
(3x+1)²-(4x+13)²=0
(3x+1+4x+13)(3x+1-4x-13)=0
(7x+14)(-x-12)=0
7x+14=0 -x-12=0
x1=-2
x2=-12