发布网友 发布时间:2024-03-07 03:42
共1个回答
热心网友 时间:2024-03-12 21:14
先化简,再求值:((x²-3/x-1)-2)÷1/x-1,其中x满足x²-2x-3=0。 原式=(x²-3-(2x-2))/(x-1)·x-1
=x²-2x-3+2
=0+2
=2
希望对你有帮助!
原式=(x^2-3/x-1-2)*(x-1)
=x^2*(x-1)-3/x-1*(x-1)+(-2)*(x-1)
=x^3-x^2-3-2x+2
=x^3-x^2-2x-1
x^2-2x-3=0两别同乘以x,x^3-2x^2-3x=0
所以原式=x^2+x-1=(2x+3)+x-1=3x+2
因为x^2-2x-3=0,所以x=-1,x=3
代入化简后的式子,得原式=-1或11
解:
原式=[(x-1)/x-(x-2)/(x+1)]÷[(2x²-x)/(x²+2x+1)]
={(x-1)(x+1)/[x(x+1)]-x(x-2)/[x(x+1)]}÷[x(2x-1)/(x+1)²]
={(x²-1-x²+2x)/[x(x+1)]}×{(x+1)²/[x(2x-1)]}
={(2x-1)/[x(x+1)]}×{(x+1)²/[x(2x-1)]}
=(x+1)/x²
∵x²-x-1=0
∴x²=x+1
∴原式=x²/x²=1
x²+2x+3(x²-2/3x)
=x²+2x+3x²-2x
=4x²
=4×¼
=1
x²(x-1)-x(x²+x-1)
=x³-x²-x³-x²+x
=-2x²+x
当x=1/2时
原式=-1/2+1/2=0
先化简,再求值:(1+x+1/1)÷x²-1/1-(x-2),其中x=-3/2
您好:
(1+x+1/1)÷x²-1/1-(x-2)
=x²-1+x-1-x+2
=x²
=(-3/2)²
=9/4
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3(x²+1/3y)-2(3/2x²-xy)+(-y+3x),
=3x²+y-3x²+2xy-y+3x
=2xy+3x
当x=-2,y=1/2时
原式=2×(-2)×1/2+3×(-2)=-2-6=-8
解:【(x²/x-1)-(2x/x-1)】÷x/x-1
=x(x-2)/(x-1)*(x-1)/x
=x-2
当:x=1/3时,
x-2=1/3-2=-5/3 .