câu hỏi? 

Tìm min

b, ta có a³+ b³ = (a+b)(a²-ab +b²)

= (a+b)(a² -ab +b² -ab +ab)

= (a+b) ( a²-2ab +b +ab)

=(a+b) [ (a²-b²) +ab ]

vậy ...........................

câu a bạn sai đề à

đề c/ m : ( a+b)(a²-ab +b²) +(a-b)(a²+ab+b² ) =2a³

(a+ b)(a² -ab +b²) + (a-b)(a²+ab +b²)

= a³+b³+a³-b³(hdt)

= 2a³

^{chúc bạn học tốt}

\(\left(a+b\right)\left(a^2-ab+b^2\right)-\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a^3+b^3\right)-\left(a^3-b^3\right)=2b^3\left(đpcm\right)\)

tik mik nha

giúp vs mn ơi

\(\left(a+b\right)\left(a^2-ab+b^2\right)-\left(a-b\right)\left(a^2+ab+b^2\right)\)

\(=a^3+b^3-a^3+b^3\)

\(=2b^3\)

1/

\(\left(1\right)=\left(a^3+b^3\right)+\left(a^3-b^3\right)=2a^3\)

2/

\(\left(2\right)=a^3+b^3=\left(a+b\right).\left(a^2-ab+b^2\right)\)

\(\left(2\right)=\left(a+b\right).\left[\left(a^2-2ab+b^2\right)+ab\right]=\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)

3/

\(\left(3\right)=\left(ac\right)^2+\left(ad\right)^2+\left(bc\right)^2+\left(bd\right)^2\)

\(\left(3\right)=\left[\left(ac\right)^2+2acbd+\left(bd\right)^2\right]+\left[\left(ad\right)^2-2adbc+\left(bc\right)^2\right]\)(do t/c giao hoán trong phép nhân => 2acbd=2adbc)

\(\left(3\right)=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

\(\sum\dfrac{a}{b^2+bc+c^2}\ge\dfrac{\left(a+b+c\right)^2}{ab^2+abc+ac^2+bc^2+abc+ba^2+ca^2+abc+cb^2}=\dfrac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}=\dfrac{a+b+c}{ab+bc+ac}\)

Đúng rầu đấy

(a-b)² = (a-b).(a-b)

         = a² - ab - ab + b²

         = a² - 2ab + b² (đpcm)