Question

Phân tích đa thức thành nhân tử: \(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

 

Answer 1

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

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

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

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

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

Answer 2

Lời giải:
$4b^2c^2-(b^2+c^2-a^2)^2=(2bc)^2-(b^2+c^2-a^2)^2$

$=(2bc-b^2-c^2+a^2)(2bc+b^2+c^2-a^2)$

$=[a^2-(b^2+c^2-2bc)][(2bc+b^2+c^2)-a^2]$

$=[a^2-(b-c)^2][(b+c)^2-a^2]$

$=(a-b+c)(a+b-c)(b+c-a)(b+c+a)$