Question

[Nguyễn Quốc Khánh]

Phân tích đa thức thành nhân tử:

a) (6x-1)^2 - (3x+2)

b) 9(2x+3)^2 - 4(x+1)^2

c) 4b^2c^2 - (b^2+c^2-a^2)^2

d) (a^2 + b^2 -5)^2 - 4(ab+2)^2

Answer 1

[Kiệt Nguyễn]

a) \(\left(6x-1\right)^2-\left(3x+2\right)^2\)

\(=\left(6x-1+3x+2\right)\left(6x-1-3x-2\right)\)

\(=\left(9x+1\right)\left(3x-3\right)\)

\(=3\left(9x+1\right)\left(x-1\right)\)

b) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)

\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)

\(=\left(6x+9+2x+2\right)\left(6x+9-2x-2\right)\)

\(=\left(8x+11\right)\left(4x+7\right)\)

c) \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\)

\(=-\left[\left(b+c\right)^2-a^2\right]\left(b^2-2bc+c^2-a^2\right)\)

\(=-\left(b+c-a\right)\left(b+c+a\right)\left[\left(b-c\right)^2-a^2\right]\)

\(=-\left(b+c-a\right)\left(b+c+a\right)\left(b-c-a\right)\left(b-c+a\right)\)

Answer 2

[Kiệt Nguyễn]

d) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5\right)^2-\left(2ab+4\right)^2\)

\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-3^2\right]\)

\(=\left(a+b+1\right)\left(a+b-1\right)\left(a-b-3\right)\left(a-b+3\right)\)