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![](https://rs.olm.vn/images/avt/0.png?1311)
\(F=a^3+b^3+ab\left(a+b\right)+2a+b+\frac{3}{a}+\frac{2}{b}\)
\(F=\left(a+b\right)^3-3ab\left(a+b\right)+ab\left(a+b\right)+a+b+a+\frac{1}{a}+\frac{2}{a}+\frac{2}{b}\)
\(F=8-4ab+2+a+\frac{1}{a}+\frac{2}{a}+\frac{2}{b}\)
Ta có: \(\left(a+b\right)^2\ge4ab\Leftrightarrow-4ab\ge-\left(a+b\right)^2=-4\)
\(a+\frac{1}{a}\ge2\sqrt{a.\frac{1}{a}}=2\)
\(\frac{2}{a}+\frac{2}{b}\ge\frac{8}{a+b}=4\)
Suy ra \(F\ge8-4+2+2+4=12\)
Dấu \(=\)xảy ra khi \(a=b=1\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(F=\left(a+b\right)^3-3ab\left(a+b\right)+2ab+2+a+\frac{2a+3b}{ab}\)
\(=8-6ab+2ab+2+a+\frac{b+4}{ab}\)
\(=10-4ab+a+\frac{1}{a}+\frac{4}{ab}\)
\(F\ge10-\left(a+b\right)^2+2\sqrt{\frac{a}{a}}+\frac{4}{\frac{1}{4}\left(a+b\right)^2}=12\)
\(F_{min}=12\) khi \(a=b=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. (a2+b2+ab)2-a2b2-b2c2-c2a2
=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2
=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2
=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)
=(a2+b2)[(a+b)2-c2]
=(a2+b2)(a+b+c)(a+b-c)
2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2
3. a(b3-c3)+b(c3-a3)+c(a3-b3)
=ab3-ac3+bc3-ba3+ca3-cb3
=a3(c-b)+b3(a-c)+c3(b-a)
=a3(c-b)-b3(c-a)+c3(b-a)
=a3(c-b)-b3(c-b+b-a)+c3(b-a)
=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)
=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)
=(a-b)(c-b)(a2+ab+2b2+bc+c2)
4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)
5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]
=2b(3a2+b2)
6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]
=(x-y-1)(x2+y2+xy-2x-y+1)
7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)
(Đúng nhớ like nhá !)
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