用因式分解法解方程①(2x-1)²-x²-4x-4=0 ②x²-2007x-2008=0③x²-(根号2+根号3)x+根号6=0 ④(x-3)²+2X(x-3)=0 ⑤(x-1)(x+3)=12
问题描述:
用因式分解法解方程①(2x-1)²-x²-4x-4=0 ②x²-2007x-2008=0
③x²-(根号2+根号3)x+根号6=0 ④(x-3)²+2X(x-3)=0 ⑤(x-1)(x+3)=12
答
①(2x-1)²-x²-4x-4=0(2x-1)²-(x+2)²=0(2x-1+x+2)(2x-1-x-2)=03x+1=0或x-3=0得x=-1/3或x=3②x²-2007x-2008=0(x+1)(x-2008)=0x+1=0或x-2008=0得x=-1或x=2008③x²-(根号2+根号3)x+根号...