X2+1/X2

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X2+1/X2
x2-3x-1=0,求①x2 1/x2;②x2-1/x2

x2-3x-1=0,求①x21/x2;②x2-1/x2x2-3x-1=0,求①x21/x2;②x2-1/x2x2-3x-1=0,求①x21/x2;②x2-1/x2∵x2-3x-1=0∴X-2-1/X=0∴X-1/X=3∴①平方,X²

x2-3x+1=0 x2/(x4-x2+1)

x2-3x+1=0x2/(x4-x2+1)x2-3x+1=0x2/(x4-x2+1)x2-3x+1=0x2/(x4-x2+1)请问这是什么啊?希望能把问题完善了好吧?

(X2 -y+1)(X2+1)+X2y+y -X2因式分解

(X2-y+1)(X2+1)+X2y+y-X2因式分解(X2-y+1)(X2+1)+X2y+y-X2因式分解(X2-y+1)(X2+1)+X2y+y-X2因式分解(X^2-y+1)(X^2+1)+X^2y+y-X^2=(X^2-y+1)(X

x2-5x+1=0则x2+x2/1

x2-5x+1=0则x2+x2/1x2-5x+1=0则x2+x2/1x2-5x+1=0则x2+x2/1你可以参见“韦达定理”方程两个根的积是1,说明他们互为倒数.x^2+1/x^2=(x+1/x)^2-2*x*1/x=(-5)²-

7/x2+x+1/x2-x=6/x2-1

7/x2+x+1/x2-x=6/x2-17/x2+x+1/x2-x=6/x2-17/x2+x+1/x2-x=6/x2-17/x(x+1)+1/x(x-1)=6/(x+1)(x-1)两边乘x(x+1)(x-1)7x-7+x+1=6x2x=6x

1+1x2-1x2-6=?

1+1x2-1x2-6=?1+1x2-1x2-6=?1+1x2-1x2-6=?-6

X1+1/X1-X2-1/X2 因式分解

X1+1/X1-X2-1/X2因式分解X1+1/X1-X2-1/X2因式分解X1+1/X1-X2-1/X2因式分解X1+1/X1-X2-1/X2=(X1-X2)+(1/X1-1/X2)=(X1-X2)(1-1/(x1x2))

因式分解(x2+1)y2-x2-1

因式分解(x2+1)y2-x2-1因式分解(x2+1)y2-x2-1因式分解(x2+1)y2-x2-1(x2+1)y2-x2-1=x2y2+y2-x2-1=(x2y2-x2)+(y2-1)=x2(y2-1)+(y2-1)=(x2-1)(y2

求arctanx/(x2(1+x2))的不定积分?

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∫arctanx/x2(1+x2)dx答案?

∫arctanx/x2(1+x2)dx答案?∫arctanx/x2(1+x2)dx答案?∫arctanx/x2(1+x2)dx答案?看图片

lim(根号X2+X-根号X2+1)

lim(根号X2+X-根号X2+1)lim(根号X2+X-根号X2+1)lim(根号X2+X-根号X2+1)x趋近无穷?如果是无穷,答案是1/2先有理化,然后再分子分母各除以x

dx/(x2根号(x2-1))不定积分

dx/(x2根号(x2-1))不定积分dx/(x2根号(x2-1))不定积分dx/(x2根号(x2-1))不定积分取x=sect(t在第一象限)原式=∫costdt=sint+C=1/sqrt(1-1/x^2)+C若t在第二象限原式=-∫c

计算(x3+x2+x+2)/(x2+1)

计算(x3+x2+x+2)/(x2+1)计算(x3+x2+x+2)/(x2+1)计算(x3+x2+x+2)/(x2+1)(x3+x2+x+2)/(x2+1)=(x3+x+x2+1+1)/(x2+1)=x+1+1/(x2+1)【OK?】请给出

4x2-(x2+1)2因式分解

4x2-(x2+1)2因式分解4x2-(x2+1)2因式分解4x2-(x2+1)2因式分解4x^2-x^4-2x^2-1-x^4+2x^2-1-(x^2-1)^2-(x+1)^2(x-1)^24x²-(x²+1)

6x2-[-3x2-(x-1)] 化简

6x2-[-3x2-(x-1)]化简6x2-[-3x2-(x-1)]化简6x2-[-3x2-(x-1)]化简6x2-[-3x2-(x-1)]=6x2+3x2+x-1=9x2+x-1=6x²-(-3x²-x+1)=6x&#

x2-1/x2-5x+6

x2-1/x2-5x+6x2-1/x2-5x+6x2-1/x2-5x+6是解不等式:(x^2-1)/(x^2-5x+6)

求不定积分x2(x2+1)分之一

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x2次方+x/x2次方-1,分式

x2次方+x/x2次方-1,分式x2次方+x/x2次方-1,分式x2次方+x/x2次方-1,分式(x²+x)/(x²-1)=[x(x+1)]/[(x+1)(x-1)]=x/(x-1)

约分 1-x2/x2-3x-4

约分1-x2/x2-3x-4约分1-x2/x2-3x-4约分1-x2/x2-3x-41-x2=(1-x)(1+x)x2-3x-4=(x+1)(x-4)共同约去(x+1)即可

(1-x2)(x2-4x+5)

(1-x2)(x2-4x+5)(1-x2)(x2-4x+5)(1-x2)(x2-4x+5)因为(x2-4x+5)是永远大于零的,它的b2-4ac