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M = (a + b + c)3 - a3 - b3 - c3
= (a + b)3 + c3 + 3(a + b)2c + 3(a + b)c2 - a3 - b3 - c3
= a3 + b3 + c3 + 3a2b + 3ab2 + 3(a + b)c(a + b + c) - a3 - b3 - c3
= 3ab (a + b) + 3c(a + b)(a + b + c)
= 3(a + b)[ab + c(a + b + c)]
= 3(a + b)(ab + bc + ac + c2)
= 3(a + b)[b(a + c) + c(a + c)]
= 3(a + b)(b + c)(c + a)
N = a3 + b3 + c3 - 3abc
= (a + b)3 + c3 - 3ab(a + b) - 3abc
= (a + b + c)3 - 3(a + b)c(a + b + c) - 3ab(a + b + c)
= (a + b + c)[(a + b + c)2 - 3(a + b)c - 3ab]
= (a + b + c)(a2 + b2 + c2 + 2ab + 2bc + 2ca - 2ac - 3bc - 3ab)
= (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
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Ta có: \(D=ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+3abc\)
\(=a^2b+ab^2+b^2c+bc^2+ac^2+a^2c+3abc\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
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Co P=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)
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sửa đề thành \(ab\left(a+b\right)+bc\left(b+c\right)+ca\left(c+a\right)+2abc\)
\(=ab\left(a+b\right)+b^2c+bc^2+c^2a+ca^2+2abc\)
\(=ab\left(a+b\right)+\left(b^2c+abc\right)+\left(c^2a+c^2b\right)+\left(a^2c+abc\right)\)
\(=ab\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ac\left(a+b\right)\)
\(=\left(a+b\right)\left(ab+bc+a^2+ca\right)\)
\(=\left(a+b\right)\left[\left(ab+bc\right)+\left(c^2+ac\right)\right]\)
\(=\left(a+b\right)\left[b\left(a+c\right)+c\left(c+a\right)\right]\)
\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ab\left(a+b\right)-bc\left(b+c\right)+ca\left(a+c\right)+abc\)
\(=a^2b+ab^2-b^2c-bc^2+ca^2+c^2b+abc\)
\(=a^2b+ab^2-b^2c+a^2c+abc\)
Đến đây thì mk chịu
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ab(a+b) + bc(b+c) + ca(c+a) = a^2b + ab^2 + b^2c + bc^2 + ca(c+a) + 2abc
= ab^2 + b^2c + a^2b + bc^2 + 2abc + ca(c+a)
=b^2(a+c) + b(a^2 + c^2 + 2ac) + ca(c+a)
=b^2(a+c) + b(a+c)^2 + ca(c+a)
=(c+a)[b^2 + b(a+c) + ca]
=(c+a)[b^2 + ab + bc + ca]
=(c+a)[b(b+a) + c(b+a)]
=(c+a)(b+c)(b+a)
![](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)
\(A=\left(a+b+c\right).\left(bc+ca+ab\right)-abc\\ =abc+b^2c+bc^2+a^2c+abc+ac^2+a^2b+ab^2+abc-abc\\ =\left(b^2c+bc^2\right)+\left(a^2b+a^2c\right)+\left(ac^2+abc\right)+\left(ab^2+abc\right)\\ =bc\left(b+c\right)+a^2\left(b+c\right)+ac\left(b+c\right)+ab\left(b+c\right)\\ =\left(b+c\right)\left(bc+a^2+ac+ab\right)\\ =\left(b+c\right)\left[a\left(a+b\right)+c\left(a+b\right)\right]=\left(b+c\right)\left(a+c\right)\left(a+b\right)\)
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a: Ta có: \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)
\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)
\(=\left[\left(a-b\right)^2-9\right]\cdot\left[\left(a+b\right)^2-1\right]\)
\(=\left(a-b-3\right)\left(a-b+3\right)\left(a+b-1\right)\left(a+b+1\right)\)