【分式的加减】几道计算题(带过程)我验算一下
(1)(1 / x)-( 1 / x+1)=[(x+1)-x]/(x+1)x=x/(x²+x)
(2)(x / x-2)- (8 / x ² -4)=[x(x+2)-8]/(x ² -4)=[(x-2)(x+4)]/[(x-2)(x+2)]=(x+4)/(x+2)
(3)(x / x+1)- (1 / x-1)- (2 / x ² -1)=[x(x-1)-(x+1)-2]/(x ² -1)=(x ²-2x-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 / x ² -3x+2)- (x+2 / x ² +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/(x ²-x-2)
(6)【(x ² -4 / x ² -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)
人气:228 ℃ 时间:2020-04-03 02:41:41
解答
1)(1 / x)-( 1 / x+1)=[(x+1)-x]/(x+1)x=x/(x²+x)
(2)(x / x-2)- (8 / x ² -4)=[x(x+2)-8]/(x ² -4)=[(x-2)(x+4)]/[(x-2)(x+2)]=(x+4)/(x+2)
(3)(x / x+1)- (1 / x-1)- (2 / x ² -1)=[x(x-1)-(x+1)-2]/(x ² -1)=(x ²-2x-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 / x ² -3x+2)- (x+2 / x ² +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/(x ²-x-2)
(6)【(x ² -4 / x ² -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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