Question

Phân tích các đa thức sau thành nhân tử bằng phương pháp dùng hằng đẳng thức:
1) \(16x^2-9\)
2) \(x^2-4+\left(x+2\right)^2\)
3) \(5a\left(a-2\right)-a+2\)
4) \(7\left(a-5\right)+8a\left(5-a\right)\)
5) \(25a^2-4b^2+4b-1\)

Answer 1

1, \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right)\left(4x+3\right)\)

2,\(x^2-4+\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)\left(x+2\right)^2=\left(x-2\right)\left(x+2\right)^3\)

3,\(5a\left(a-2\right)-a+2=5a\left(a-2\right)-1\left(a-2\right)=\left(5a-1\right)\left(a-2\right)\)

4,\(7\left(a-5\right)+8a\left(5-a\right)=7\left(a-5\right)-8a\left(a-5\right)=\left(7-8a\right)\left(a-5\right)\)

5, \(25a^2-4b^2+4b-1=25a^2-\left(4b^2-4b+1\right)=\left(5a\right)^2-\left(2b-1\right)^2=\left(5a-2b+1\right)\left(5a+2b-1\right)\)

Answer 2

1) (4x-3)(4x+3)

2) (x-2)(x+2)+(x+2)² = (x+2)(x-2+x+2) = 2x(x+2)

3) 5a(a-2)-(a-2) = (a-2)(5a-1)

4) 7(a-5)-8a(a-5) = (a-5)(7-8a)

5) (25a²-1)+(-4b²+4b) = (5a-1)(5a+1)-4b(b-1)