Câu hỏi của lê sỹ phát

Question

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

lê thị hương giang · Accepted Answer

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

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

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

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

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

$=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)$Câu trả lời: $4b^2c^2-\left(b^2+c^2-a^2\right)^2$

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

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

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

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

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

Ôn tập: Phân thức đại số