Bài 6:

a) \(a^2+b^2+1\ge ab+a+b\) ⇔ \(2\left(a^2+b^2+1\right)\ge2\left(ab+a+b\right)\)

⇔ \(2a^2+2b^2+2\ge2ab+2a+2b\) ⇔ \(\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\) ≥ 0

⇔\(\left(a+b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\)≥ 0 luôn đúng

b) \(a^3+b^3\ge ab^2+a^2b\) ⇔ \(\left(a+b\right)\left(a^2-ab+b^2\right)\ge ab\left(a+b\right)\)

⇔\(\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)⇔\(\left(a+b\right)\left(a^2-ab+b^2-ab\right)\)

⇔\(\left(a+b\right)\left(a^2-2ab+b^2\right)\)⇔\(\left(a+b\right)\left(a-b\right)^2\)≥0 luôn đúng với a,b >0

xin lx vì phần c mik chưa nghĩ ra

cô nói thế thì chịu rồi :))

Studying hard for exam leads to student's tiredness

cái đấy hỏi cô chứ hỏi mik chi z tr

1. C

2. D

New devices have been invented to help people live a longer life.

B

B

A

B

A

B