Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=ab\left(a+b\right)-bc\left(b+a\right)-bc\left(c-a\right)-ac\left(c-a\right)\)
\(=\left(a+b\right)\left(ab-bc\right)+\left(a-c\right)\left(bc-ac\right)\)
\(=\left(a+b\right)\cdot b\left(a-c\right)+\left(a-c\right)\cdot c\left(b-a\right)\)
\(=\left(a-c\right)\left(ab+b^2+cb-ac\right)\)
b: \(=ab^2+ac^2+bc^2+a^2b+a^2c+b^2c+2abc\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2\)
\(=\left(a+b\right)\left(ab+c^2+ac+cb\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
d: \(=a^3\left(b-c\right)-b^3\left(b-c+a-b\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-b^2-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\cdot\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
a(b2+c2)+b(c2+a2)+c(a2+b2)+22abc
= ab2+ac2+bc2+a2b+(a2c+b2c+2abc)
= ab(a+b)+c2(a+b)+c(a+b)2
= (a+b)(ab+c2+ac+bc)
= (a+b)[a(b+c)+c(b+c)
= (a+b)(b+c)(a+c)
b)
(a+b)(a2-b2)+(b+c)(b2-c2)+(a+c)(c2-a2)
= (a+b)(a2-b2)-(b+c)[(a2-b2)+(c2-a2)] +(a+c)(c2-a2)
= (a2-b2)(a+b-b-c) +(c2-a2)(a+c-b-c)
= (a2-b2)(a-c)+(c2-a2)(a-b)
= (a-b)(a2-ac+ab-bc +c2-a2)
= (a-b)[a(b-c)-c(b-c)]
= (a-b)(b-c)(a-c)
\(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\\ =a^3\left(c-b^2\right)+ab^3-b^3c^2+bc^3-a^2c^3+a^2b^2c^2-abc\\ =a^3\left(c-b^2\right)+bc^2\left(c-b^2\right)-ab\left(c-b^2\right)-a^2c^2\left(c-b^2\right)\\ =\left(c-b^2\right)\left(a^3+bc^2-ab-a^2c^2\right)\\ =\left(c-b^2\right)\left[a^2\left(a-c^2\right)-b\left(a-c^2\right)\right]\\ =\left(c-b^2\right)\left(a-c^2\right)\left(a^2-b\right)\)
a)= ab (a + b) - bc [( a + b) - (a - c)] + ac (a - c)
= ab (a + b) - bc (a + b) + bc (a - c) +ac (a - c)
= b (a + b) (a - c) + c (a - c) (a + b)
= (a + b) (a - c) (b + c)
b) \(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left[\left(a^2-b^2\right)+\left(c^2-a^2\right)\right]+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a^2-b^2\right)\left[\left(a+b\right)-\left(b+c\right)\right]+\left(c^2-a^2\right)\left[\left(c+a\right)-\left(b+c\right)\right]\)
\(=\left(a-b\right)\left(a+b\right)\left(a-c\right)-\left(a-c\right)\left(a+c\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\left[\left(a+b\right)-\left(a+c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
1) \(\left(a-b\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)
\(a\left(b+c\right)^2\left(b-c\right)+b\left(c+a\right)^2\left(c-2\right)+c\left(a+b\right)^2\left(a-b\right)\)
\(=\left(b-c\right)\left(c-a\right)\left(c-b\right)\left(c+b+a\right)\)
nguồn câu hỏi tương tự
Trang 136 trong nâng cao phát triển có viết rồi mình cóp nó vô để mọi người dễ đọc nhé !
