在实数范围内分解因式:(1)x눀-2x-6 (2)2x눀+8x-7 (3)x눀-5xy+3y눀

请用解方程的方法做,写清楚过程!O(∩_∩)O谢谢~
2025-03-04 04:15:05
推荐回答(2个)
回答1:

(1)令x^2-2x-6=0,用求根公式解得两根为x1=-2,x2=4
所以,x²-2x-6=(x+2)(x-4)
(2)的方法同一:结果是2[x-(4-根号30)/2][x-(4+根号30)/2]
(3)将y看成常数解方程可得结果为[x-(5-根号13/2)y][x-(5+根号13/2)y]

回答2:

(1)x²-2x-6

=(x²-2x+1)-7
=(x-1)²-(√7)²
=(x-1+√7)(x-1-√7)

(2)2x²+8x-7

=(2x²+8x+8)-15
=2(x²+4x+4)-15
=2(x+2)²-15
=[√2(x+2)]²-(√15)²
=(√2x+2√2+√15)(√2x+2√2-√15)

(3)x²-5xy+3y²

=[x²-2×5/2xy+(5/2y)²]-13/4y²
=(x-5/2y)²-(√13/2y)²
=(x-5/2y+√13/2y)(x-5/2y-√13/2y)
=1/2×(2x-5y+√13y)×1/2×(2x-5y-√13y)
=1/4[2x-(5-√13)y][2x-(5+√13)y]

不懂的还可以追问!满意请及时采纳! O(∩_∩)O