设X-X分之1=2,求X²加X²分之1的值

(x^2-3x+1)/(x^2-4)=30/(x+2)(x^2-3x+1)/(x-2)=30x^2-3x+1=30x-60x^2-33x+61=0x=(33±√1333)/2

(x^2-3x+1)/(x^2-4)=30/(x+2)(x^2-3x+1)/(x-2)=30x^2-3x+1=30x-60x^2-33x+61=0x=(33±√1333)/2

你这个题目稍微有点乱啊……如果我看的不错的话应该是这样——（5x+2)/(x^2+x)=3/(x+1)其中x不等于0和-1交叉相乘得（5x+2)(x+1)=3(x^2+x)整理得2x^2+4x+2=0

（x*x+1）/x=x+1/x因为根号x+根号1/x=3,所以等式两边同时平方得x+1/x+2=9,所以x+1/x=9-2=7

答：1）x/(x^2-x)+2=2x/(x+1)1/(x-1)=2x/(x+1)-2=-2/(x+1)-2x+2=x+13x=1x=1/3经检验,x=1/3是原分式方程的解2）x/(x+1)-1=3/

f(x)=根号下x的平方加1分之x平方加3x?是f(x)=√[(x²+3x)/(x²+1)]吗?如果是的话：f(x)=√[(x²+3x)/(x²+1)]令：x=

x^2-3x+1=0x+1/x=(x^2+1)/x=(x^2-3x+1+3x)/x=(0+3x)/x=3x^2+1/x^2=(x+1/x)^2-2=3^2-2=7

1/X(X+1)+1/(x+1)(x+2)+.+1/(x+99)(x+100)=1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+99)-1/(x+100)=1/x-1/(x+10

你现在高一么?如果是,那就应该是印刷问题应该是4的y次方吧x=lg36/lg3,y=lg36/lg42/x+1/y=(2lg3+lg4)/lg36=1

(x²+2)/[x(x+1)(x+2)]=A/x+B/(x+1)+C/(x+2)(x²+2)/[x(x+1)(x+2)]=[A(x+1)(x+2)]+Bx(x+2)+Cx(x+1)

f(-1)=(-1)^3=-1,f(0)=0,f(1)=e^(-1)=1/e定义域为R.【分段函数定义域为各段自变量取值范围之并】

f(x)=(2x^2+2x+1)/(x+1)=(2x^2+4x+2-2x-2+1)/(x+1)=[2(x+1)^2+1-2(x+1)]/(x+1)=2(x+1)+1/(x+1)-2因为x>-1,x+1

2/(x+3)(x-3）+(x-2)/x(x-3)=1/x(x+3)两边乘x(x+3)(x-3)2x+x^2+x-6=x-3x^2+2x-3=0(x+3)(x-1)=0x+3在分母,不等于0所以x=1

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

f(x)+2f(1/x)=x用1/x代替x得：f(1/x)+2f(x)=1/x两边同时乘2得：2f(1/x)+4f(x)=2/x和原式相减得：3f(x)=2/x-x所以f(x)=2/(3x)-x/3

x+2分之2x减2=x-2分之1加4-x分之x+14左右两边同×（x+2）(x-2)(4-x)得：（2x-2）(x-2)(4-x)=(x+2)(4-x)+(x+14)(x+2)(x-2)2(x-2)&

x/(x^2-x+1)=7,(x^2-x+1)/x=1/7x+(1/x)-1=1/7x+(1/x)=8/7x^2/(x^4+x^2+1)=1/[(x^4+x^2+1)/x^2]=1/[x^2+(1/x

每个分式减去1(x-4-x+5)/(x-5)+(x-8-x+9)/(x-9)=(x-7-x+8)/(x-8)+(x-5-x+6)/(x-6)1/(x-5)+1/(x-9)=1/(x-8)+1/(x-6

1+1/(x+2)+1+1/(x+3)=1+1/(x+4)+1+1/(x+1)1/(x+2)+1/(x+3)=1/(x+4)+1/(x+1)1/(x+3)-1/(x+4)=1/(x+1)-1/(x+2

x(x+3)分之1是可以把它们分解成x分之1-（x+3）分之1然后总体在乘以3分之1的,后面的每个式子都可以分解成两个分式相减然后再乘以3分之1的,由于式子很多中间的一些是可以抵消的,最后就只有剩下几