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

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

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

b: \(a^m-a^{m+2}\)

\(=a^m-a^m\cdot a^2\)

\(=a^m\left(1-a^2\right)\)

\(=a^m\left(1-a\right)\left(1+a\right)\)

\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)

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

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

\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

\(=\left(a+b-1\right)\left(a+b+1\right)\left(a-b-3\right)\left(a-b+3\right)\)

a)

(ax+by)² - (ay+bx)²

=(ax+by-ay-bx)(ax+by+ay+bx)

=[ a(x-y) -b(x-y)][ a(x+y) + b(x+y)]

=(a-b)(x-y)(a+b)(x+y)

b)(a²+b²-5)² - 4(ab+2)²

=(a²+b²-5-2ab-4)(a²+b²-5+2ab+4)

=[ (a-b)² -9][ (a+b)² -1]

=(a-b-3)(a-b+3)(a+b-1)(a+b+1)

a, \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+ay+bx+by\right)\left(ax-ay+bx-by\right)\)

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

b, \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)

\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b+1\right)\left(a+b-1\right)\)

\(a\left(b^2+c^2\right)+b\left(a^2+c^2\right)+c\left(a^2+b^2\right)+2abc\)

\(=ab^2+ac^2+ba^2+bc^2+ca^2+cb^2+2abc\)

\(=\left(ab^2+ba^2\right)+\left(ac^2+bc^2\right)+\left(ca^2+abc\right)+\left(cb^2+abc\right)\)

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

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

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

hình như cộng 2abc chứ sao +2ab

Trong đề kt 1t trên lớp mình làm ghi có 2ab à. Vậy chắc là đề sai nhỉ. 

Thanks nha

\(5x.\left(x-10\right)-2x+2x+20\)

\(=5x^2-50x+20\)

\(=5\left(x^2-10x+5^2-21\right)\)

\(=5\left[\left(x-5\right)^2-\left(\sqrt{21}\right)^2\right]\)

\(=5\left(x-5-\sqrt{21}\right)\left(x-5+\sqrt{21}\right)\)

\(a\left(a-b\right)^2\left(a+b\right)-\left(b-a\right)^2\left(a^2-5ab+b^2\right)\)

\(=a\left(a-b\right)^2\left(a+b\right)-\left(a-b\right)^2\left(a^2-5ab+b^2\right)\)

\(=\left(a-b\right)^2\left[a.\left(a+b\right)-a^2+5ab-b^2\right]\)

\(=\left(a-b\right)^2\left[a^2+ab-a^2+5ab-b^2\right]\)

\(=\left(a-b\right)^2\left(6ab-b^2\right)\)

Sửa đề: \(\left(a-b\right)^2-\left(b-a\right)\left(a-3b\right)\)

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

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

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2.\left(a-b\right)\left(a-2b\right)\)

Tham khảo nhé~

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

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

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

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

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

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

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

Chúc bạn học tốt.

→(a+b)(a²-b²) +(b+c)(b²-a²) -(c²-a²)(b+c) +(c+a)(c²-a²)

↔(a²-b²)(a+b-b-c)-(c²-a²)(b+c-c-a)

↔(a-c)(a²-b²)-(c²-a²)(b-a)

↔(a-c)((a²-b²-(a+c)(b-a))

↔(a-c)(a-b)(a+b+b-a)

↔2b(a-c)(a-b)