\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)

\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)\)

\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a\right)\)

\(=\left(a+b+c\right)\left(ab+bc\right)+ca\left(c+a\right)\)

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

\(=\left(a+c\right)\left[b.\left(a+b+c\right)+ca\right]\)

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

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

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

\(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)

\(=ab\left(a+b\right)+abc+bc\left(b+c\right)+abc+ca\left(c+a\right)+abc\)

\(=ab\left(a+b+c\right)+bc\left(b+c+a\right)+ca\left(c+a+b\right)\)

\(=\left(a+b+c\right)\left(ab+bc+ac\right)\)

Tham khảo nhé~

thank you

ko bt đâu thông cảm

phân tích bằng đặt ẩn phụ=))

Ta có:\(\left(a^2+b^2+c^2\right)\left(a+b+c\right)^2+\left(ab+bc+ca\right)^2\)

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

Đặt:\(a^2+b^2+c^2=x;ab+bc+ca=y\),ta có:

\(x\left(x+2y\right)+y^2=x^2+2xy+y^2=\left(x+y\right)^2\)

Thay vào,ta được:\(\left(x+y\right)^2=\left(a^2+b^2+c^2+ab+bc+ca\right)^2\)

wtf ko bt ma cung dí

(x^2-x+2)^2+(x-2)^2 
= [(x^2-x+2)+(x-2)]^2-2[(x^2-x+2)*(x-2)] (áp dụng (a^2+b^2)=(a+b)^2-2ab 
=(x^2)^2- 2((x^3-3x^2+4x-4) 
=x^4-2x^3+6x^2-8x+8 
 giờ phân tích đa thức 
x^4-2x^3+6x^2+8x-8 
=(x^4-2x^3+2x^2)+(4x^2-8x+8) (cái này làm bài tập nhiêu nhìn ra nhanh) 
=[x^2(x^2-2x+2)]+4(x^2-2x+2) dẹp luôn 
=(x^2-2x+2)(x^2+4) 

\(\left(x^2-x+2\right)^2+\left(x-2\right)^2\)

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

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

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

\(=\left(x-2\right)^2\left(x^2+2x+2\right)\)

Cảm ơn hai bạn nha nhưng Đường Quỳnh Giang nhầm dấu rồi.😆

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

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

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

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

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

Ta có b + c = (a + b) + (c – a) nên

A = ab(a + b) – bc[(a + b) + (c – a)] – ac(c – a)

= ab(a + b) – bc(a + b) – bc(c – a) – ac(c – a)

= b(a + b)(a – c) – c(c – a)(b + a)

= (a + b)(a – c)(b + c)

Đáp án cần chọn là: B

ab(a-b) + bc((b-a)+(a-c)) +ac(c-a) 
=ab(a-b) -bc(a-b) -bc(c-a) +ac(c-a) 
=(a-b)(ab-bc) +(c-a)(ac-bc) 
=(a-b) b (a-c) + (c-a) c (a-b) 
=(a-b)(a-c)(b-c)