化简(x²-4X+4分之x²-4+x+2分之2-x)除以x-2分之x 求解
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是的,刚刚答错了题了
[(x²-4x+4)/(x²-4)+(x+2)/(2-x)]/[(x-2)/x]
=[(x-2)/(x+2)+(2-x)/(x+2)]/[x/(x-2)]
=0/[x/(x-2)]
=0
[(x²-4)/(x²-4x+4)+(2-x)/(x+2)]/[x/(x-2)]
=[(x+2)/(x-2)+(2-x)/(x+2)]/[x/(x-2)]
=[x+2+(2-x)(x-2)/(x+2)]/x
=8/(x+2)
不清楚谁是分子分母。
[(x²-4x+4)/(x²-4)+(x+2)/(2-x)]/[(x-2)/x]
=[(x-2)/(x+2)+(2-x)/(x+2)]/[x/(x-2)]
=0/[x/(x-2)]
=0
[(x²-4)/(x²-4x+4)+(2-x)/(x+2)]/[x/(x-2)]
=[(x+2)/(x-2)+(2-x)/(x+2)]/[x/(x-2)]
=[x+2+(2-x)(x-2)/(x+2)]/x
=8/(x+2)
不清楚谁是分子分母。
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(x²-4X+4分之x²-4+x+2分之2-x)除以x-2分之x
=[(x-2)(x+2)/(x-2)²+(2-x)/(x+2)]【x-2】/x
=[(x+2)/(x-2)+(2-x)/(x+2)]*(x-2)/x
=[(x+2)²-(x-2)²]/(x+2)(x-2)*(x-2)/x
=8x/(x+2)*1/x
=8/(x+2)
=[(x-2)(x+2)/(x-2)²+(2-x)/(x+2)]【x-2】/x
=[(x+2)/(x-2)+(2-x)/(x+2)]*(x-2)/x
=[(x+2)²-(x-2)²]/(x+2)(x-2)*(x-2)/x
=8x/(x+2)*1/x
=8/(x+2)
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