已知x²+x—1=0,求(x+1)(x²—2x+6)-(-x)²+(2x-1)²的值 速求
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![已知x²+x—1=0,求(x+1)(x²—2x+6)-(-x)²+(2x-1)²的值 速求](/uploads/image/z/1786143-39-3.jpg?t=%E5%B7%B2%E7%9F%A5x%26%23178%3B%2Bx%E2%80%941%3D0%2C%E6%B1%82%28x%2B1%29%28x%26%23178%3B%E2%80%942x%2B6%29-%28-x%29%26%23178%3B%2B%282x-1%29%26%23178%3B%E7%9A%84%E5%80%BC+%E9%80%9F%E6%B1%82)
已知x²+x—1=0,求(x+1)(x²—2x+6)-(-x)²+(2x-1)²的值 速求
已知x²+x—1=0,求(x+1)(x²—2x+6)-(-x)²+(2x-1)²的值 速求
已知x²+x—1=0,求(x+1)(x²—2x+6)-(-x)²+(2x-1)²的值 速求
已知x²+x—1=0,那么:x²=-x+1
所以:
(x+1)(x²—2x+6)-(-x)²+(2x-1)²
=(x+1)(-x+1-2x+6)-x²+4x²-4x+1
=(x+1)(-3x+7) +3x²-4x+1
=-3x²+4x+7+3x²-4x+1
=8