Question

Phân tích đa thức sau thành nhân tử:

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

b, \(16xy+4y^2-9+16x^2\)

Answer 1

\(a,4a^2b^2-\left(a^2+b^2-1\right)^2\)
\(=\left(2ab\right)^2-\left(a^2+b^2-1\right)^2\)
\(=\left[2ab-\left(a^2+b^2-1\right)\right]\left[2ab+\left(a^2+b^2-1\right)\right]\)
\(=\left(2ab-a^2-b^2-1\right)\left(2ab+a^2+b^2-1\right)\)
\(=\left[-\left(a^2+2ab+b^2\right)-1\right]\left[\left(a^2+2ab+b^2\right)-1\right]\)
\(=\left[-\left(a+b\right)^2-1^2\right]\left[\left(a+b\right)^2-1^2\right]\)
\(=\left[\left(-a-b-1\right)\left(-a-b+1\right)\right]\left[\left(a+b-1\right)\left(a+b+1\right)\right]\)
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\(b,16xy+4y^2-9+16x^2\)
\(=\left(4y^2+16xy+16x^2\right)-9\)
\(=\left(2y+4x\right)^2-3^2\)
\(=\left(2y+4x-3\right)\left(2y+4x+3\right)\)