1 x-1=1 2x-2

是这样的：x^5+x^4=x^3(x^2+x)=(x^2+x)[(x^3-1)+1]=(x^2+x)(x^3-1)+x^2+x=[x(x+1)(x-1)](x^2+x+1)+x^2+x=(x^3-x)

1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x+13x+14x+15x=550120x=550x=55/12=4.583

2x(x-1)-x(3x+2)=-x(x+2)-122x^2-2x-3x^2-2x=-x^2-2x-122x^2-3x^2+x^2-2x-2x+2x=-12-2x+12=0-2(x-6)=0x-6=0

(x+2-12/x-2)/(4-x)/(x-2),其中x=-1/2=(x²-4-12)(x-2)÷(4-x)/(x-2)=(x²-16)/(4-x)=-(4-x)(4+x)/(4-

(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=(x/2)*(2/x(x+2))+(2/(x+2)(x+4))+.+2/(x+8)(x+10)=(x/2)*[1/x

对不等式变形有[x-x^2+7x-12]/(x^2-7x+12)>=0,又因为x^2-7x+12=(x-3)(x-4)-x^2+8x-12=-(x-2)(x-6),于是有-(x-2)(x-3)(x-4

x/(x2+1)=(x-1)/xx3-2x2+x-1=0设y=x3-2x2+x-1y=x2(x-2)+x-2+1=(x-2)(x2+1)+1x增,y也增,所以y是单增函数,图像与x轴有交点,即f（x）

这样算,分离变量:x²+x=(x+1)²-(x+1)然后,除下来,就等于x+1-1=x注意,x≠-1!

(1)12x+36-6x=102则：6x=66所以：x=11(2)2x-48-x=54,即：x=54+48=102(3)2x-1+x=14,即：3x=15,所以：x=5

每个绝对值里都是正数所以原式=x+1+x+3+x+5+……+x+13-x-2-x-4-x-6-……-x-12=x-1*6+13=x+7=π/2+7

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

=9x^4+4(x+1)-3x^2-7x+1=9x^4-3x^2-3x+5=9x^4-3x(x+1)+5=9x^4-3x×3x^3+5=5

60X+15X+10X+6X=12091X=120X=120/91

x/2+x/6+x/12+x/20+x/30+x/42=136x/42=1x=7/6

应该是-2x²,不是-3x²3x²-x=1原式=9x^4-3x³+15x³-5x²+3x²-7x+2001=3x²(3x

x/x(x-1)+2=2x/(x+1)1/(x-1)+2=2x/(x+1)x+1+2(x-1)(x+1)=2x(x-1)x+1+2x²-2-2x²-2x=0-x-1=0-x=1x=

1/6x+1/12x+1/7x+5+1/2x+4=x2/12x+1/12x+6/12x-12/12x+1/7x=-9-1/4x+1/7x=-9-7x+4x=-9*28-3x=-9*28x=84

=x^2002(x^4+x^3+x^2+x+1)=0提取公因式就行了

答：结论是无解的设1和4中间的正方形边长为x则左边中间的正方形边长为x+1左下角边长为x+1+x=2x+1所以：右下角正方形边长2x+1+x-4=3x-3所以：最大的正方形底部边长=2x+1+3x-3

方程右边的1减到左边来,再同分得到（6x-2)/(x^2-x)=0得到x=1/3这个方程即使两边乘以x^2-x,x的最高次数为2次,也不会出现x^3,你肯定是什么地方弄错了