120x*6x2=900

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120x*6x2=900
解方程:1/x2+x +1/x2+3x+2 +1/x2+5x+6 +1/x2+7x+12 +1/x2+9x+20=5/x

后面的x²+11x-708有误吧!再问:没有题目就这样能不能帮我再答:那我就试试:原式为:1/x2+x+1/x2+3x+2+1/x2+5x+6+1/x2+7x+12+1/x2+9x+20=5

解方程 2/(x2-x)+6/(1-x2)=7/(x2+x)

2/(x2-x)+6/(1-x2)=7/(x2+x)2/x(x-1)-6/(x-1)(x+1)=7/x(x+1)[x*(x-1)*(x+1)]*[2/x(x-1)-6/(x-1)(x+1)]=[7/x

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

解分式方程:1/(x2-2x-3) +2/(x2-x-6) +3/(x2+3x+2)=0

原式可化为1/(x+1)(x-3)+2/(x-3)((x+2)+3/(x+1)(x+2)=0两边同乘以:(x+1)(x+2)(x-3)得:(x+2)+2(x+1)+3(x-3)=0(x≠-1,x≠-2

x2-5x+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)²-2=23

解分式方程:2/x2+5x+6 + 3/x2+x-6=4/x2-4

等式两边同时乘以(x+3)(x-2)(x+2)就可以去分母了

方程7 / x2+x+ 3/ x2-x=6 / x2-x怎么解?

7/(X2+X)+3/(X2-X)=6/(X2-X),去分母,等式两端同时乘X(X+1)(X-1):7(X-1)+3(X+1)=6(X+1),7X-7+3X+3=6X+6,7X+3X-6X=6+7-3

先化简,再求值:9x+6x2-3(x-23x2),其中x=-1.

原式=9x+6x2-3x+2x2=8x2+6x,当x=-1时,原式=8×(-1)2+6×(-1)=8-6=2.

x2+5x+6/2+4-x2/4+x2+x-6/3=0 解方程,检验!

2x+5x+3+4-x/2+2x+x-2=019x/2=-5x=-19/10检验:带入x是、原式=0

解方程7/(x+x2)-3/(x-x2)=6/(x2-1)

7/(x+x2)-3/(x-x2)=6/(x2-1)两边同乘以x(x+1)(x-1),得7(x-1)+3(x+1)=6x7x-7+3x+3=6x10x-6x=3-74x=-4x=-1经检验x=-1是增

解方程:x2+1+x=6/x+x2

令x²+x=t原方程变为t+1=6/tt²+t-6=0(t+3)(t-2)=0则t=2或-31)x²+x=2x²+x-2=0(x+2)(x-1)=0x=-2或x

1/(x2+x)+1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)=4/21

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

x · 根号(X2+3X+18) - X · 根号(X2-6X+18)=1 那么2X · 根号(X2-6X+18)-9X

因为X*根号(X^2+3X+18)-X*根号(X^2-6X+18)=1则X*根号(X^2+3X+18)=X*根号(X^2-6X+18)+1两边平方得X^2*(X^2+3X+18)=1+X^2*(X^2

解方程 √X2+6X+2-√X2+X-2=X

x=0.问下那个根号x2是不是根号x的平方的意思啊.再问:是的,有没有过程再答:有啊。√X²+6x+2-√X²+x-2=x→7x=x故x=0.(化简)

用换元法解(2(x2+1)/x)+(6x/x2+1)=7其中(x2+1/x)=y

就把(x2+1/x)=y代进去(2(x2+1)/x)+(6x/x2+1)=7化为2y+6/y=7解之得y=3/2或2(x2+1/x)=3/2时无解(x2+1/x)=2时x=1再问:用换元法把(x2+1

已知x2+x-6=0,求代数式x2(x+1)-x(x2-1)-7的值.

x2(x+1)-x(x2-1)-7,=x3+x2-x3+x-7,=x2+x-7,∵x2+x-6=0,∴x2+x-7=-1,即x2(x+1)-x(x2-1)-7=-1.

x1+x2=8 /x1-x2/ x /c/=6 x1,x2,c 为整数

不怎么看的懂,写清楚!

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-

已知x2+y2+4x-6y+13=0 求x2-2x\x2+3y2

即(x²+4x+4)+(y²-6y+9)=0(x+2)²+(y-3)²=0所以x+2=y-3=0x=-2,y=3所以原式=(4+4)/(4+27)=8/31

求函数f(x)=x2/x2-4x+1(x≥6)的值域

答:f(x)=x²/(x²-4x+1),x>=6分子分母同除以x²得:f(x)=1/(1-4/x+1/x²)=1/[(1/x-2)²-3]因为:x>=