一元二次方程49x42255x12的解移项[5(5x-1)]2-[7(x-4)]2=0[5(5x-1)+7(x-4)][5(5x-1)-7(x-4)]=0(32x-33)(18x+23)=0x=33/32，x=-23/18x2-5x+7=1x2-5x+6=0(x-2)(x-3)=0x1=2x2=32，12223242526210021012用简便方法计算12-22+32-42+52-62+……-1002+1012=1+(3-2)*(3+2)+(5-4)*(5+4)+...+(101+100)*(101