解√3/3x+3=2/3x^2-√3x+1

一个“整式的乘法”的问题请先阅读下列解题过程,再仿做下面的问题.已知X*X + X - 1=0,求X*X*X + 2*X*X + 3的值.解X*X*X + 2X*X +3=X*X*X +X*X -X +X*X +X +3=X{X*X +X -1} +X*X +X -1 +4=0+0+4=4+ x + X*X + X*X*X=0.+ X*X + X*X*X 已知x*x-3x+1=0求√(x*x+1/x-2)=? 解方程x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6) 解方程4x^2+2x√(3x^+x)+x-9=0 *-----------------------------------------------*| 6 4 X | 8 X X | X X 5 || X X X | X X X | X 7 8 || X X X | X X X | X X X ||---------------+---------------+--------------- || X X X | X X X | 5 1 X || X X X | X 6 X | X X X || 8 X X | 3 5 X | 2 X X || x(x*x-6x-9)-x(x*x-8x-15)+2x(3-x) 填九宫格帮帮忙.x x 6 x x 7 x x 98 x x x 3 x 1 x x 9 x x 6 x 5 x 3 x x x 3 x x x x 1 8x x x 9 x 1 x x x2 1 x x x x 6 x x x 6 x 7 x 3 x x 1 x x 9 x 2 x x x 47 x x 8 x x 5 x x xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 七年级下册政治复习提纲(山东人民出版社)第五单元是 青春的脚步 青春的气息别弄错了啊!格式：X X X X X X X X X 1、X X X X X X X X X X.2、X X X X X X X X X.3、X X X X X X X X X X X. (X+1)X(X+2)X(X+3)=504 先化简在求值(x-1/3x-x+1/x)*x/x²-1,其中x=√2-3 求值域:f(x)=|x-1|+|x+3|+|x-2|f(x)=x+√(2x-1) x(x-2)(x-3) {x|(x+3)(x-2) -x^3+x^2-x 解方程】(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3)(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3) 解方程】(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3)(x+8)/(x-3)-(x-9)/(x-4)=(x+7)/(x+8)-(x+2)/(x+3) (x*x-3)*(x*x-3)-x*(x*x-3)-2x*x 因式分解