4x2-27x+18>0 x2-6x+4

4x2-27x+18>0 x2-6x+4 解一元二次不等式组 4x2-27x+18>0 x2-6x+8 解下列不等式组 4x2-27x+18>0 x2+4x+4>0 (x2-x)(x2-x-4)+4因式分解 计算:x2-5x+6/x2-4x+3 + x2-4x+3/x2-2x+1 + x2-3x-4/1-x2 4x2-4x+1=x2+6x+9 (x2+6x)(x2+6x+4)-45 (x2+5x-4)(x2+5x+6)-120 因式分解 (x2+2x-3)(x2+2x+4)+6因式分解 已知x2 3x-2＝0,求2x-4╱x2 6x 9÷（x-3）×x2-9╱2x-x2的值 因式分解x2-4x-6 已知x2－4x+1=0,求x2／(x2)2－3x2+1 几道数学题,解不等式X2-4X-5≤0-2X2+3X-2＜0-X2+8X-3＞04X2-18X+81/4＞01/2X2-3X+5＜03X2-2X+1＜0X2+2X-3≤0X-X2+6＜04X2+4X+1≥0-4+X-X2＜03X2-4＜02X-X2≥-14-X2≤0（X-A）（X-2）＜0 x2+x-6=0 x2-根号3x-二分之一=0 4x2-6x=0 解方程4x2-4x+1=x2+6x+9 7x2-√6x-5=0 x2-4x+1=0,则x+1/x=?x2+1/x2=? x2-6x+4≥0 x2-4x+4