Câu hỏi của Phác Chí Mẫn

Question

Phân tích đa thức thành nhân tử:

$\left(1+x^2\right)^2-4x\left(1-x^2\right)$

Nguyễn Thành Trương · Accepted Answer

\left(1+x^2ight)^2-4x\left(1-x^2ight)\

=\left(1+x^2ight)^2-4x\left(1-x^2ight)+4x^2-4x^2\

=\left(1-x^2-2xight)^2-4x^2\

=\left(1-x^2-2x+2xight)\left(1-x^2-2x-2xight)\

=\left(1-x^2ight)\left(1-x^2-4xight)\

=\left(1-xight)\left(1+xight)\left(1-x^2-4xight)Câu trả lời: \left(1+x^2ight)^2-4x\left(1-x^2ight)\

=\left(1+x^2ight)^2-4x\left(1-x^2ight)+4x^2-4x^2\

=\left(1-x^2-2xight)^2-4x^2\

=\left(1-x^2-2x+2xight)\left(1-x^2-2x-2xight)\

=\left(1-x^2ight)\left(1-x^2-4xight)\

=\left(1-xight)\left(1+xight)\left(1-x^2-4xight)

Đan Anh · Answer

$\left(1+x^2\right)^2-4x\left(1-x^2\right)$

$=\left[\left(1+x^2\right)-4x^2\right]+4x\left(1-x^2\right)+4x^2$

$=\left(1+x^2-2x\right)\left(1+x^2+2x\right)+4x\left(1-x^2\right)+4x^2$

$=\left(1-x\right)^2\left(1+x\right)^2-4x\left(1-x^2\right)+4x^2$

$=\left(1-x^2\right)^2-4x\left(1-x^2\right)+4x^2$

$=\left(1-x^2-2x\right)^2$Câu trả lời: $\left(1+x^2\right)^2-4x\left(1-x^2\right)$

$=\left[\left(1+x^2\right)-4x^2\right]+4x\left(1-x^2\right)+4x^2$

$=\left(1+x^2-2x\right)\left(1+x^2+2x\right)+4x\left(1-x^2\right)+4x^2$

$=\left(1-x\right)^2\left(1+x\right)^2-4x\left(1-x^2\right)+4x^2$

$=\left(1-x^2\right)^2-4x\left(1-x^2\right)+4x^2$

$=\left(1-x^2-2x\right)^2$

Violympic toán 8