Question

求几道分解因式的答案（需过程）

1、3x^2+19xy+6y^2
2、ax^n+1-ax^n-20ax^n-1
3、(1-x^2)(1-y^2)+4xy

Answer 1

1、3x^2+19xy+6y^2 3 1
=(3x+y)(x+6y)---->用交叉相乘法 1 6
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2.、ax^n+1-ax^n-20ax^n-1
=ax^(n-1)(x^2-x-20)
=ax^(n-1)(x-5)(x+4)

3、(1-x^2)(1-y^2)+4xy
=1 - X^2 - Y^2 + X^2 * Y^2+ 4XY
=( 1 + 2XY + X^2 * Y^2 ) - ( X^2 - 2XY + Y^2 )
=( 1 + XY )^2 - ( X - Y )^2
=( 1 + XY + X - Y )( 1+ XY - X + Y )

Answer 2

1、3x^2+19xy+6y^2
=3x^2+18xy+xy+6y^2
=3x(x+6y)+y(x+6y)
=(x+6y)(3x+y)

2、ax^n+1-ax^n-20ax^n-1
=ax^(n-1)(x^2-x-20)
=ax^(n-1)(x-5)(x+4)

3、(1-x^2)(1-y^2)+4xy
=x^2y^2-x^2-y^2+1+2xy+2xy
= (x^2y^2+2xy+1)-(x^2-2xy+y^2)
=(xy+1)^2-(x-y)^2
=(xy+x-y+1)(xy-x+y+1)

Answer 3

(1)
3x² + 19xy + 6y²
= (3x + y)(x + 6y)

(2)
ax^(n + 1) - ax^n - 20ax^(n - 1)
= ax^(n - 1)·(x² - x - 20)
= ax^(n - 1)·(x + 4)(x - 5)

(3)
(1 - x²)(1 - y²) + 4xy
= x²y² - x² - y² + 1 + 4xy
= (x²y² + 2xy + 1) - (x² + y² - 2xy)
= (xy + 1)² - (x - y)²
= (xy + 1 + x - y)(xy + 1 - x + y)

Answer 4

1、（3x+y）（x+6y）