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(x+1)(x-2)

来源:baiyundou.net   日期:2024-08-19

谷费瑶3332x - 2的绝对值+x+1的绝对值的最小值 -
宇舒胃18144027729 ______[答案] |X-2|+|X+1| 当X2时.原式=X-2+X+1=2X-1 随X的增大而增大,当X=2时最小为3 所以原式的最小值为3.

谷费瑶3332解方程:2/x+1 - 1/x - 2=0 -
宇舒胃18144027729 ______ 解: 2/(x+1)-1/(x-2)=0 2(x-2) -(x+1) =0 2x-4 -x-1=0 x-5=0 x=5

谷费瑶3332比较(x 1)(x - 2)与(x +3)(x - 4)的大小, -
宇舒胃18144027729 ______[答案] 做差法: (x+3)(x-2)-2(x-4) =x²+x-6-2x+8 =x²-x+2 =(x-1/2)²+7/4>0 故(x+3)(x-2)>2(x-4)

谷费瑶3332先化简再求值x+2分之1 - x+2分之x的平方+2x+1除以x - 1分之x的平方 - 1其中x=根号2 - 2 -
宇舒胃18144027729 ______ 原式=[1/(x+2)]-[(x+2x+1)/(x+2)]÷[(x-1)/(x-1)] =[1/(x+2)]-[(x+1)/(x+2)]÷[(x+1)(x-1)/(x-1)] =[1/(x+2)]-[(x+1)/(x+2)]÷(x+1) =[1/(x+2)]-[(x+1)/(x+2)]*[1/(x+1)] =[1/(x+2)]-[(x+1)/(x+2)] =(1-x-1)/(x+2) =-x/(x+2) 当x=√2-2时 原式=-(√2-2)/(√2-2+2) =(2-√2)/√2 =√2(2-√2)/2 =(2√2-2)/2 =√2-1 希望满意!

谷费瑶3332化简求值 X/X+2 - X2+2X+1/X+2 ÷ X2 - 1/X - 2 其中X=√3 - 2 -
宇舒胃18144027729 ______ X/X+2 - X2+2X+1/X+2 ÷ X2-1/X-2 =[x/(x+2)]-[(x+1)^2/(x+2)]*{(x-2)/[(x+1)(x-1)]}=[x/(x+2)]-{(x+1)(x-2)/[(x+2)*(x-1)]}={1/(x+2)}*{x-[(x^2-x-2)/(x-1)]}={1/(x+2)}*{2/(x-1)}=2/[(x+2)(x-1)]=2/[√3*(√3-3)]=-(1+√3)/3 √希望你能看懂,你能明白, 望采纳,赞同

谷费瑶3332x^3 - 3x - 2化简结果是(x+1)^2*(x - 2) -
宇舒胃18144027729 ______[答案] 解 x³-3x-2 =x³-x-2x-2 =x(x²-1)-2(x+1) =x(x-1)(x+1)-2(x+1) =(x+1)[x(x-1)-2] =(x+1)(x²-x-2) =(x+1)(x-2)(x+1) =(x+1)²(x-2)

谷费瑶33322(x+1) - (x - 2)=3(4 - x),(y - 2) - 2(y+2)=3(y+4)+6,2(2x+1)=(5x+5) -
宇舒胃18144027729 ______ 2(x+1)-(x-2)=3(4-x) 2x+2-x+2=12-3x 4x=8 x=2 (y-2)-2(y+2)=3(y+4)+6 y-2-2y-4=3y+12+6 4y=-24 y=-6 2(2x+1)=5x+5 4x+2=5x+5 x=-3

谷费瑶3332解方程|x - 2|+|x+1|=6 -
宇舒胃18144027729 ______ 解:①当x<-1时,|x-2|+|x+1|=-(x-2)-(x+1)=-2x+1=6,解得x=-5/2②当-1≤x≤2时,|x-2|+|x+1|=-(x-2)+(x+1)=3≠6,不符合,舍去③当x>2时,|x-2|+|x+1|=x-2+x+1=2x-1=6,解得x=7/2综上

谷费瑶3332已知f(x+1)=X^2 - 3x+2 求f(x) -
宇舒胃18144027729 ______[答案] f(x+1)=x^2-3x+2=(x-1)(x-2)=[(x+1)-2][(x+1)-3],所以f(x)=(x-2)(x-3).

谷费瑶33323(x+1) - 3/1(x - 1)=2(x - 1) - 2/1(x+1) -
宇舒胃18144027729 ______ 3(x+1)-3/1(x-1)=2(x-1)-2/1(x+1) 两边乘618(x+1)-2(x-1)=12(x-1)-3(x+1)18x+18-2x+2=12x-12-3x-316x+20=9x-1516x-9x=-20-157x=-35 x=-5 不明白,可以追问如有帮助,记得采纳 如追加其它问题,采纳本题后另发并点击向我求助,谢谢 祝学习进步!

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