x平方减ax 1=0和x平方-x-a=0有一个公共根

x²-5x+1=0除以x得x-5+1/x=0x+1/x=5x²+1/x²=(x+1/x)²-2=5²-2=23

3x^2-7x+3=0等式两边同除x得3x-7+3/x=0即x+1/x=7/33x^2-7x+3=0（1）等式两边同除x^2得3-7/x+3/x^2=0（2）（1）,（2）两式相加得3x^2+3/x^

0＜y＜x【(x^2+y^2)(x-y)】-【(x^2-y^2)(x+y)】=【(x^2+y^2)(x-y)】-【(x-y)(x+y)^2】=(x-y)【(x^2+y^2-(x+y)^2】=(x-y)

x^2+3x+1=0x+1/x=-32x^2+2/x^2=2[(x+1/x)^2]-4=2*9-4=14

x^2-3x+1=0x^2+1=3x同时除以xx+1/x=3所以x^2+1/x^2=(x+1/x)^2-2=3^2-2=7x^2/(x^4+x^2+1)=1/(x^2+1+1/x^2)=1/8

x^2-3x+1=0x^2+1=3xx+(1/x)=(x^2+1)/x=3x/x=3x^2+[1/(x^2)]=[x+(1/x)]^2-2x*(1/x)=3^2-2=7

x平方-x+1=x平方-x分之6-x+1=-x分之6x-x分之6-1=0x平方-x-6=0(x-3)(x+2)=0x-3=0,x+2=0所以,x=3.x=-2

x²-3x+1=0x²+1=3x同时除以xx+1/x=3(x+1/x)²=9x²+1/x²+2=9x²+1/x²=7x²

X平方减3X+1=0,方程两边除以x,得x-3+1/x=0所以x+1/x=3X平方+X平方分之1=x²+2+1/x²-2=(x+1/x)²-2=3²-2=7

X平方=-3x-1第一题的答案就出来了啊X平方=2004-a=2005-b=2006-ca=b-1=c-2a/bc=a/(a+1)(a+2)后面的都一样,把b和c全换成带有a的式子,最后应该能算出答案

X平方+3X+1=0两边同除xx+1/x=-3(x+1/x)^2=9x^2+1/x^2+2=9x^2+1/x^2=7

(x^2-2x)^2-5(x^2-2x)-6=(x^2-2x-6)(x^2-2x+1)=(x^2-2x-6)(x-1)^2=0所以x-1=0或者x^2-2x-6=0x^2-2x-6=0x^2-2x+1

(x平方+x)平方-3x平方-3x+2=0(x²+x)²-3(x²+x)+2=0令x²+x=A,则原方程可化为A²-3A+2=0(A-1)(A-2)=

解:令x²-x=y则原方程可化为y²+y-6=0(y+3)(y-2)=0解得y=-3或y=2当y=-3时,x²-x=-3即x²-x+3=0判别式△=(-1)&#

1+x+x平方+x立方+x的4次方+.+x的2008次方+x的2009次方=(1+x+x²)+x³(1+x+x²)+.+x^2007(1+x+x²)=0+0+.

设y=x²+x原方程即y(y-2)=-1y²-2y+1=0(y-1)²=0y-1=0y=1x²+x-1=0x=(-1±√5)/2

x^2-2=0x^2=2原式=x^2-2x+1+x^2/x+2=2-2x+1+2/x+2=(2x+4-2x^2-4x+x+2+2)/x+2=(2x+4-4-4x+x+2+2)/(x+2)=4-x/x+

5x^2-13x+5=0x^2-13x/5+1=0x-13/5+1/x=0x+1/x=13/5(x-1/x)^2=(x+1/x)^2-4(x-1/x)^2=(13/5)^2-4(x-1/x)^2=16

A是不是=｛x|ax-1=0｝?如果是因为A是B的子集,所以A有可能是空集也有可能是B的非空子集1‘A为空集时,a=02’A为B的非空子集时,因为B={x|x^2-3x+2=0}={x|(x-2)(x

x^2+x=1x^2=1-x2x^3-3x^2-x=2x(x^2+x)-5x^2-x=2x-5(1-x)-x=6x-5.