{[2(X+1)]+4}÷2-X+517.
来源:学生作业帮助网 编辑:作业帮 时间:2024/08/24 01:48:25
原式=[(x+2)/(x^2-2x)-(x-1)/(x^2-4x+4)]÷2(x-4)/(x^3-2x^2)=[(x+2)/(x^2-2x)-(x-1)/(x-2)²)]*[x²(
[(x+2)/(x*x-2x)-(x-2)/(x*x-4x+4)]/[(x-4)/x]=[(x+2)/x(x-2)-(x-2)/(x-2)²]/[(x-4)/x]=[(x²-4-x
(x+1-1/1-x)÷(x-x²/x-1)=[(x+1)+1/(x-1)]×[-(x-1)/x(x-1)]=(x²-1+1)/(x-1)×(-1/x)=-x/(x-1)(x-4/
=(x+1)(x-1)/(x+2)²×1/(x+1)×(x+2)(x+1)/(x-1)=(x+1/(x+2)
是这样的:x^5+x^4=x^3(x^2+x)=(x^2+x)[(x^3-1)+1]=(x^2+x)(x^3-1)+x^2+x=[x(x+1)(x-1)](x^2+x+1)+x^2+x=(x^3-x)
|x-1|+|x-10|表示数轴上x到1的距离+x到10的距离.显然最小值是9,此时x只要在1到10之间就好.类似的,|x-2|+|x-9|的最小值是7,此时x在2到9之间就好.|x-3|+|x-8|
(x^2-4/x^2-4x+4)÷(x+2/x+1)-x/x-2=(x+2)/(x-2)(x+2)*(x+1)/(x+2)-x/x-2=(x+1)/(x-2)(x+2)-x(x+2)/(x-2)(x+
(x+2/x²-2x-x-1/x²-4x+4)÷x-4/x={(x+2)(x-2)/[x(x-2)²]-x(x-1)/[x(x-2)²]}·x/(x-4)={(
你写的看不懂分子分母的代数式用括号括起来否则分不出来先通分【((x-2)(x+2)-x(x-1))/x(x+2)(x+2)】(x+2)/(x-4)=(x-4)/x(x+2)(x-4)=1/x(x+2)
注意:本题x=2时题目没意义,应该舍去(这是很多学生错误所在)这里x只能取根号2-1
原式=[x(x−2)x(x−2)+2(x−2)x(x−2)-x(x+1)x(x−2)]•x(x−2)x+4=−(x+4)x(x−2)•x(x−2)x+4=-1.
解(4x²-1)/(2-4x)÷(4x²+4x+1)/x=(2x-1)(2x+1)/2(1-2x)×[x/(2x+1)²=-(2x+1)/2×[x/(2x+1)²
解1:原式=x(x-1)/(x+1)*(x+1)/x=x-1解2:原式=(x+3)^2/(x+3)(x-3)*(x-3)/(x+2)=(x+3)/(x-3)*(x-3)/(x+2)=(x+3)/(x+
2x/(x-2)(x+1)÷(1-x/x+1)-x²+2x/x²-4?=2x/(x-2)(x+1)×(x+1)/(1-x)-x(x+2)/(x+2)(x-2)=2x/(x-2)(1
.[(x+4)(x-2)]/x³+2x²+x÷[(x-2)/x·(x+4)/(x+1)]=[(x+4)*(x-2)]/[x(x+1)^2]/[(x-2)*(x+4)/(x(x+1)
[(x-2/x²+2x)-(x-1/x²+4x+4)]÷﹙x-4/x+2﹚=[(x-2/x(x+2)-(x-1/(x+2)²]÷﹙x-4/x+2﹚=[(x-2)(x+2)
原式=〖(x+2)/x(x-2)-(x-1)/(x-2)^2〗÷〖(x-4)(x+4)/x(x+4)〗=〖(x+2)(x-2)-(x-1)x〗/x(x-2)^2〗÷〖(x-4)(x+4)/x(x+4)
x(2-1/x)+x/(x^2-2x)÷(3-x)/(x^2-4)=x((2x-1)/x)+x/(x^2-2x)x(x^2-4)/(3-x)=(2x-1)+(x+2)/(3-x)=(7x-3-2x^2
/>(x+2)/(x+1)-(x+3)/(x+2)-(x+4)/(x+3)+(x+5)/(x+4)=1+1/(x+1)-1-1/(x+2)-1-1/(x+3)+1+1/(x+4)=1/(x+1)-1/
你的括号丢得太多了,我猜猜猜1、实在猜不出来,哪里该有括号2、2X/(2X-1)+X/(X-2)=2[2X(X-2)+X(2X-1)]/(2X-1)(X-2)=22X^2-4X+2X^2-X=4X^2