先化简,再计算∶x^2+2x/1+x÷(x-2/x+1),其中x=—1/2
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先化简,再计算∶x^2+2x/1+x÷(x-2/x+1),其中x=—1/2
原式=[(x²+2x)/(1+x)]÷[(x-2)/(x+1)]
=[(x²+2x)/(x+1)]×[(x+1)/(x-2)]
=(x²+2x)/(x-2)
=[(-1/2)²+2×(-1/2)]/(-1/2-2)
=(1/4-1)/(-5/2)
=(-3/4)×(-2/5)
=3/10
再问: 是这样的x减2分之x+1) 不是这样 [(x-2)/(x+1)]
再答: 你好:是这样吗? 原式=[(x²+2x)/(x+1)]÷[(x-2)/x+1] =[(x²+2x)/(x+1)]÷[(x-2)/x+(x/x)] =[(x²+2x)/(x+1)]÷[(x-2+x)/x] =[x(x+2)/(x+1)]×{x/[2(x-1)] =x²(x+2)/(x²-1) =(x³+2x²)/(x²-1) =[(-1/2)³+2×(-1/2)²]/[(-1/2)²-1] =(-1/8+1/2)/(-3/4) =3/8×(-4/3) =-1/2
=[(x²+2x)/(x+1)]×[(x+1)/(x-2)]
=(x²+2x)/(x-2)
=[(-1/2)²+2×(-1/2)]/(-1/2-2)
=(1/4-1)/(-5/2)
=(-3/4)×(-2/5)
=3/10
再问: 是这样的x减2分之x+1) 不是这样 [(x-2)/(x+1)]
再答: 你好:是这样吗? 原式=[(x²+2x)/(x+1)]÷[(x-2)/x+1] =[(x²+2x)/(x+1)]÷[(x-2)/x+(x/x)] =[(x²+2x)/(x+1)]÷[(x-2+x)/x] =[x(x+2)/(x+1)]×{x/[2(x-1)] =x²(x+2)/(x²-1) =(x³+2x²)/(x²-1) =[(-1/2)³+2×(-1/2)²]/[(-1/2)²-1] =(-1/8+1/2)/(-3/4) =3/8×(-4/3) =-1/2
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