2(x-1)2-(x+1)(x-1)-4=2(x2-2x+1)-(x2-1)-4=x2-4x-1,
∴原式=(x2-4x)-1=3-1=2.
已知x2-4x-3=0,求2(x-1)2-(x+1)(x-1)-4的值.
已知x2-4x-3=0,求2(x-1)2-(x+1)(x-1)-4的值.
數(shù)學(xué)人氣:600 ℃時(shí)間:2020-03-29 18:15:38
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