用因式分解法解方程①(2x-1)눀-x눀-4x-4=0 ②x눀-2007x-2008=0

2024-12-09 22:09:36
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回答1:

①(2x-1)²-x²-4x-4=0
(2x-1)²-(x+2)²=0
(2x-1+x+2)(2x-1-x-2)=0
3x+1=0或x-3=0
得x=-1/3或x=3

②x²-2007x-2008=0
(x+1)(x-2008)=0
x+1=0或x-2008=0
得x=-1或x=2008

③x²-(根号2+根号3)x+根号6=0
(x-根号2)(x-根号3)=0
x-根号2=0或x-根号3=0
得x=根号2或x=根号3

④(x-3)²+2X(x-3)=0
(x-3)(x-3+2x)=0
x-3=0或3x-3=0
即x=3或x=1

⑤(x-1)(x+3)=12
x²+2x-3-12=0
x²+2x-15=0
(x-3)(x+5)=0
x-3=0或x+5=0
得x=3或x=-5