问题标题:
【分式的加减】几道计算题(带过程)(1)(1/x)-(1/x+1)=(2)(x/x-2)-(8/x²-4)=(3)(x/x+1)-(1/x-1)-(2/x²-1)=(4)(1/x+1/y)²-(1/x-1/y)²=(5)(x-1/
问题描述:
【分式的加减】几道计算题(带过程)
(1)(1/x)-(1/x+1)=
(2)(x/x-2)-(8/x²-4)=
(3)(x/x+1)-(1/x-1)-(2/x²-1)=
(4)(1/x+1/y)²-(1/x-1/y)²=
(5)(x-1/x²-3x+2)-(x+2/x²+3x+2)=
(6)【(x²-4/x²-x-6)-(x+2/x-3)】÷(4/x-3)=
李亚辉回答:
(1)(1/x)-(1/x+1)=[(x+1)-x]/(x+1)x=x/(x²+x)
(2)(x/x-2)-(8/x²-4)=[x(x+2)-8]/(x²-4)=[(x-2)(x+4)]/[(x-2)(x+2)]=(x+4)/(x+2)
(3)(x/x+1)-(1/x-1)-(2/x²-1)=[x(x-1)-(x+1)-2]/(x²-1)=(x²-2x-3)/[(x-1)(x+1)]=[(x-3)(x+1)]/[(x-1)(x+1)]=(x-3)(x-1)
(4)(1/x+1/y)²-(1/x-1/y)²=[(1/x+1/y)-(1/x-1/y)][(1/x+1/y)+(1/x-1/y)]=4/xy
(5)(x-1/x²-3x+2)-(x+2/x²+3x+2)=(x-1)/(x-1)(x-2)-(x+2)/(x+1)(x+2)=1/(x-2)-1/(x+1)=[(x+1)-(x-2)]/(x-2)(x+1)=3/(x²-x-2)
(6)【(x²-4/x²-x-6)-(x+2/x-3)】÷(4/x-3)=(x-2)(x+2)/(x+2)(x-3)-(x+2)/(x-3)=(x-2)/(x-3)-(x+2)/(x-3)=[(x-2)-(x+2)]/(x-3)=4/(3-x)
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