(x/x2-9)-(1/x2+6x+9)计算

来源：学生作业帮编辑：神马作文网作业帮分类：数学作业时间：2024/11/23 23:53:50

(x/x2-9)-(1/x2+6x+9)=x/(x+3)(x-3)-1/(x+3)²
=[x(x+3)-(x-3)]/(x-3)(x+3)²
=[x²+3x-x+3]/(x-3)(x+3)²
=(x²+2x+3)/(x-3)(x+3)²

(x/x2-9)-(1/x2+6x+9)计算计算：（1）x2+9xx2+3x+x2−9x2+6x+9 (x2+3x+9)/(x2-27)+(6x)/(9x-x2)-(x-1)/(6+2x) 解方程：1/x2+x +1/x2+3x+2 +1/x2+5x+6 +1/x2+7x+12 +1/x2+9x+20=5/x x2+3x+9/x2-27+6x/9x-x2-x-1/6+2x 化简化简x2+3x+9/x2-27+6x/9x-x2-x-1/6+2x 初二数学分式计算1/(x-3)-6/(x2-9)-(x-1)/(6+2x) 2( x2 +1/x2) - 9( x + 1/x) + 14 = 化简：根号（x2+6x+9）+根号（x2-2x+1）-根号（x2-4x+4）先化简，再求值：9x+6x2-3(x-23x2)，其中x=-1．先化解,再求值：【1/(x2+3x+2)+1/(x2+5x+6)+1/(x2+7x+12)】/1/(x2+9x+20), 若根式根号(x2-6x+9)+根号(x2-4x+4)+根号(x2-10x+25)+根号(x2+2x+1)的值为常数,则实