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

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

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

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

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

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

Mình đã làm bài này rồi.

Link: https://hoc24.vn/hoi-dap/question/824554.html

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

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

\(=\left(a+b+3c\right)\left(a+b-c\right)+\left(a+b-c\right)^2\)

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

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

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

Chúc bạn học tốt :33

Thank you))))

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

\(M=ab^2-ac^2+bc^2-ba^2+c\left(a-b\right)\left(a+b\right)\)

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

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

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

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

Giờ là cách khác:(tại em làm khá kĩ nên nó dài thôi chứ em trình bày lại trong giấy nó ngắn ngủn à)

Đặt \(b^2-c^2=x;c^2-a^2=y\Rightarrow a^2-b^2=-\left(x+y\right)\)

Suy ra \(M=ax+by-c\left(x+y\right)\)

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

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

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

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

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

\(=\left(b-c\right)\left(a-c\right)\left(b-a\right)\) [muốn cho đẹp thì nhân (-1) . (-1) vào thì nó thành (a-b)(b-c)(c-a) ]