Question

x²+2x-3 2x²-7x+6 3x²y²-10xy+7 -y²+20y-96 用配方法把上述因式分解

过程详细一点 急！！！

Answer 1

x²+2x-3 =x²+2x+1-4=（x+1)^2-4 =（x+1-2）（x+1+2）=（x-1)(x+3)
3x²y²-10xy+7 =（9x²y²-30xy+25-4）/3=[（3xy-5)^2-4]/3=1/3*(3xy-7)(3xy-3)
-y²+20y-96=-(y²-20y+100-4)=-[(y-10)^2-4]=-(y-8)(y+12)

追问

2x²-7x+6 这个呢？

追答

=（16x²-56x+49-1）/8=1/8*[(4x-7)^2-1]=1/8*(4x-6)(4x-8)

Answer 2

配方法：x²+2x-3= x²+2x+1-4=(x+1)²-2²=(x+1-2)(x+1+2)=(x+1)(x+3)
十字相乘法：
2x²-7x+6=(2x-3)(x-2)
2x -3
x -2
3x²y²-10xy+7=(3xy-7)(xy-1)
3xy -7
xy -1
-y²+20y-96
= -(y²-20y+96)=-(y-12)(y-8)
y -12
y -8

Answer 3

x²+2x-3
= x²+2x+1 - 4
= (x+1)² - 4
= (x+1 + 2) (x+1 - 2)
= (x+3)(x-1)

2x²-7x+6
= 2x²-4x - 3x+6
= 2x(x-2) - 3(x-2)
= (x-2)(2x-3)

3x²y²-10xy+7
= 3x²y²-7xy -3xy+7
= xy(3xy-7) - (3xy-7)
= (3xy-7)(xy-1)

-y²+20y-96
= -y²+8y + 12y-96
= -y(y-8) + 12(y-8)
= (y-8)(-y+12)
= (y-8)(12-y)