再求值(1-x)/(x+2)²÷[x-2+3/(x+2)],其中x=1/2答案是-4/15,
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再求值
(1-x)/(x+2)²÷[x-2+3/(x+2)],其中x=1/2
答案是-4/15,
(1-x)/(x+2)²÷[x-2+3/(x+2)],其中x=1/2
答案是-4/15,
原式=(1-x)/(x+2)²÷{[(x-2)(x+2)+3]/(x+2)}
=(1-x)/(x+2)²÷[(x²-4+3)/(x+2)]
=(1-x)/(x+2)²÷[(x²-1)/(x+2)]
=(1-x)/(x+2)²×[(x+2)/(x²-1)]
=(1-x)(x+2)/[(x+2)²(x²-1)]
=-(x-1)(x+2)/[(x+2)²(x+1)(x-1)]
=-1/[(x+2)(x+1)]
=-1/(1/2+2)(1/2+1)
=-1/(15/4)
=-4/15
=(1-x)/(x+2)²÷[(x²-4+3)/(x+2)]
=(1-x)/(x+2)²÷[(x²-1)/(x+2)]
=(1-x)/(x+2)²×[(x+2)/(x²-1)]
=(1-x)(x+2)/[(x+2)²(x²-1)]
=-(x-1)(x+2)/[(x+2)²(x+1)(x-1)]
=-1/[(x+2)(x+1)]
=-1/(1/2+2)(1/2+1)
=-1/(15/4)
=-4/15
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