(x 2)(4x-1)-(2x-1)(2x 1)

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证明:根号(x2+2x+4)-根号(x2-x+1)

要使根号(x2+2x+4)-根号(x2-x+1)

(x2+1)2-4x(x2-1)因式分解 (x2-x)2+(x2+3x+2)2-4(x2+x+1)2因式分解

(x²+1)²-4x(x²-1)=(x²-1)²-4x(x²-1)+4x²=(x²-1-2x)²(x^4-2x

(x2+3x+9)/(x2-27)+(6x)/(9x-x2)-(x-1)/(6+2x)

原式=(x²+3x+9)/(x-3)(x²+3x+9)-6x/x(x-3)(x+3)-(x-1)/2(x+3)=1/(x-3)-6/(x-3)(x+3)-(x-1)/2(x+3)=

1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=1/(x+4)

1/(x²+3x+2)=[(x+2)-(x+1)]/(x+1)(x+2)=1/(x+1)-1/(x+2)同理1/(x²+5x+6)=1/(x+2)-1/(x+3)1/(x²

化简计算~1/(x2-5x+6) - 1/(4x-x2-3) - 1/(3x-x2-2)

1/(x2-5x+6)-1/(4x-x2-3)-1/(3x-x2-2)=1/(x2-5x+6)+1/(x2-4x+3)+1/(x2-3x+2)=1/(x-2)(x-3)+1/(x-3)(x-1)+1/

化简x-1分之x+1-x2-1分之x2-2x÷x2-x-2/x2+2x+1

原式=(x+1)/(x-1)-x(x-2)/(x+1)(x-1)÷(x-2)(x+1)/(x+1)²=(x+1)/(x-1)-x/(x-1)=(x+1-x)/(x-1)=1/(x-1)请好评

化简:根号(x2+6x+9)+根号(x2-2x+1)-根号(x2-4x+4)

√(x2+6x+9)+√(x2-2x+1)-√(x2-4x+4)=√(x+3)²+√(x-1)²-√(x-2)²=|x+3|+|x-1|-|x-2|①当x≤-3时,原式=

(x2+1)2-4x(x2-1)因式分解

(x^2+1)^2-4x(x^2-1)=(x^2-1)^2-4x(x^2-1)+4x^2=[x^2-1-2x]^2

因式分解 (x2-2x-x)(x2-2x+4)+9

因式分解(x2-2x-2)(x2-2x+4)+9设x^2-2x=y原式=(y-2)(y+4)+9=y^2+2y-8+9=y^2+2y+1=(y+1)^2=(x^2-2x+1)^2=(x-1)^4

(x+2)(x2-2x+4)+(x-1)(x2+x+1),其中x=-三分之二

原式=x3+8+x3-1=2x3+7=-16/27+7=173/7

解方程2/x2+x+3/x2-x=4/x2-1

两边乘x(x+1)(x-1)2(x-1)+3(x+1)=4x2x-2+3x+3=4x5x+1=4xx=-1经检验,x=-1时分母x+1=0增根,舍去方程无解

解方程 x(x-6)+2x(x-3)=3(x2-x-1) 解不等式:2x2(x-3)+4(x2-x)≥x(2x2-2x+

解方程:x2-6x+2x2-6x=3x2-3x-3-9x=-3x=1/3解不等式:2x3-6x2+4x2-4x≥2x3-2x2+5x-39x-3≤0x≤1/3

先化简,再求值:(x+2x2-2x-x-1x2-4x+4)÷x2-16x2+4x,其中x=2+3.

原式=[x+2x(x-2)-x-1(x-2)2]÷x2-16x2+4x=[x2-4x(x-2)2-x2-xx(x-2)2]÷x2-16x2+4x=x-4x(x-2)2•x(x+4)(x+4)(x-4)

先化简,在求值:(X分之x2+4-4)除以x2+2x分之x2-4,其中x是方程x2+2x+1=10的解

x²+2x+1=10(x+1)²=10x+1=3或x+1=-3所以x=2或x=-4【(x²+4)/x-4】÷【(x²-4)/(x²+2x)】=【(x&

已知x/﹙x2-3x+1﹚=2,则x2/x的4次方+x2+1=

x/(x²-3x+1)=2(x²-3x+1)/x=1/2翻过来x+1/x=7/2(x^4+x2+1)/x²翻过来=x²+2+1/x²=(x+1/x)&

x2-4x+4分之x+8 -2-x分之1 ÷ x2-2x 分之x+3

发照片呀再问:请原谅手机不行再问:请帮我一下,谢谢。再答:看不明白再问:你写出来就行了的再答: 再答:是这样吗再问:嗯,麻烦把过程写详细点再答:我说题目对不再问:是的再问:对的再问:还有一个

求多项式:3x2+4x-2x2-x+x2-3x-1的值,其中x=-3.

原式=2x2-1,当x=-3时,原式=2×(-3)2-1=17.

解方程x-4/x2+x-2=1/(x-1)+(x-6)/(x2-4)

(x-4)/(x²+x-2)=1/(x-1)+(x-6)/(x²-4)(x-4)/(x-1)(x+2)=1/(x-1)+(x-6)/(x-2)(x+2)(x-4)(x-2)=(x-

(x2+5x+4)2+2(x2+5x+4)2+1=(x2+.

解题思路:这个是因式分解问题。由完全平方公式,再应用换元法可以得到结果.解题过程:

分式计算:2x/x2-x-2÷(1-x/x+1)-x2+2x/x2-4

原式=2x/[(x-2)(x+1)]*(x+1)/(x-1)-x(x+2)/[(x+2)(x-2)]=2x/[(x-2)(x-1)]-x/(x-2)=[2x-x(x-1)]/[(x-2)(x-1)]=