![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3a^2c^2+bd+3abc+acd\)
\(=\left(3a^2c^2+3abc\right)+\left(acd+bd\right)\)
\(=3ac\left(ac+b\right)+d\left(ac+b\right)\)
\(=\left(ac+b\right)\left(3ac+d\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 25 - x2 + 4xy - 4y2 = 25 - (x2 - 4xy + 4y2) = 52 - (x - 2y)2 = (5 + x - 2y)(5 - x +2y) = (x - 2y + 5)(2y - x + 5)
b) 3a2c2 + bd + 3abc + acd = (3a2c2 + 3abc) + (bd + acd) = 3ac(ac + b) + d (ac + b) = (ac + b)(3ac + d)
c) x3 - 2x2 - x + 2 = x2(x - 2) - (x - 2) = (x - 2)(x2 - 1) = (x - 2)(x - 1)(x + 1)
d) a4 + 5a3 + 15a - 9 = (a4 + 3a2) + (5a3 + 15a) - (3a2 + 9) = a2(a2 + 3) + 5a(a2 + 3) - 3(a2 + 3) = (a2 + 3)(a2 + 5a - 3)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có :
Biến đổi vế trái thành:
a^3+b^3+c^3-3abc
<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc
<=>[(a+b)^3 +c^3] -3ab.(a+b+c)
<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c)
<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab
<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
a3+b3+c3-3abc
=(a+b)3-3ab(a+b)+c3-3abc
=[(a+b)3+c3]-3ab(a+b+c)
=(a+b+c)[(a+b)2-c(a+b)+c2]-3ab(a+b+c)
=(a=b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c)
=(a+b+c)(a2+2ab+b2-ac-bc+c2-3ab)
=(a+b+c)(a2+b2+c2-ab-ac-bc)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3a^2c^2+bd+3abc+acd=\left(3a^2c^2+3abc\right)+\left(bd+acd\right)=3ac\left(ac+b\right)+d\left(b+ac\right)\\ =\left(3ac+d\right)\left(ac+b\right)\)
\(B=a^2c-a^2d-b^2d+b^2c=a^2\left(c-d\right)-b^2\left(c-d\right)=\left(a^2-b^2\right)\left(c-d\right)\\=\left(a-b\right)\left(a+b\right)\left(c-d\right)\)
\(C=8x^2+4xy-2ax-ay=\left(8x^2+4xy\right)-\left(2ax+ay\right)=4x\left(2x+y\right)-a\left(2x+y\right)\\ =\left(4x-a\right)\left(2x+y\right)\)
\(E=3a^2-6ab+3b^2-12c^2=3\left(a^2-2ab+b^2\right)-12c^2=3\left(a-b\right)^2-12c^2\\ =3\left[\left(a-b\right)^2-4c^2\right]=3\left(a-b-2c\right)\left(a-b+2c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta Có : \(a^3+b^3+c^3-3abc \)
\(=\left(a+b\right)\left(a^2+ab+b^2\right)+c\left(c^2-3ab\right)\)
\(=a^3+3a^2b+ab^2+b^3+c^3-3abc-3a^2b-3ab^2\)
\(=\left(a+b\right)^3+c^3-3abc\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
= \(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2-ab+b^2-ac-bc+c^2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có: ab(a + b) + bc(b + c) + ac(a + c) + 3abc
= ab(a + b) + abc + bc(b + c) + abc + ac(a + c) + abc
= ab(a + b + c) + bc(a + b + c) + ac(a + b + c)
= (a + b + c)(ab + bc + ca)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3+b^3-c^3+3abc\)
\(=a^3+3ab.\left(a+b\right)+b^3-c^3-3abc-3ab.\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab.\left(a+b-c\right)\)
\(=\left(a+b+c\right).\left(a^2+ab+b^2-ab-ac+c^2\right)-3ab.\left(a+b+c\right)\)
\(=\left(a+b+c\right).\left(a^2+b^2+c^2-ab-bc-ca\right)\)
3a2c2+bd+3abc+acd=3ac(ac+b)+d(ac+b)=(3ac+d)(ac+b)