a, \(a^4+b^4-a^3b-ab^3=a^3\left(a-b\right)-b^3\left(a-b\right)\)

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

Mà \(\hept{\begin{cases}\left(a-b\right)^2\ge0\forall a;b\\a^2+ab+b^2=\left(a+\frac{1}{2}b\right)^2+\frac{3}{4}b^2\ge0\forall a;b\end{cases}}\)

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

\(\Rightarrow a^4+b^4-a^3b-ab^3\ge0\Leftrightarrow a^4+b^4\ge a^3b+ab^3\)

Dấu "=" xảy ra khi a = b

b, \(a^3-3a^2+4a+1=a\left(a^2-4a+4\right)+a^2+1=a\left(a-2\right)^2+a^2+1>0\left(\forall a>0\right)\)

c, \(a^4+b^2+2-4ab=\left(a^4-2a^2b^2+b^4\right)+\left(2a^2b^2-4ab+2\right)\)

\(=\left(a^2-b^2\right)^2+2\left(ab-1\right)^2\ge0\)

\(\Rightarrow a^4+b^4+2\ge4ab\)

Dấu "=" xảy ra khi \(\orbr{\begin{cases}a=b=1\\a=b=-1\end{cases}}\)

thank you nhá

nhân ra ik

ko rảnh mak nhân 

cái này dùng cô si

tui vừa tra mạng òi

Gọi quãng đường AB là x (x > 0 )

Do ô tô đi từ A đến B với vận tốc 30 km/h

\(\Rightarrow\)Thời gian ô tô đi từ A đến B là : \(\frac{x}{30}\)

Do ô tô đi từ B về A với vận tốc 40 km/h

\(\Rightarrow\)Thời gian ô tô đi từ  B về A là : \(\frac{x}{40}\)

     \(\text{Đ}\text{ổi}\)15 phút =  \(\frac{1}{4}gi\text{ờ}\)

               2h30' = \(\frac{5}{2}\)giờ

Do ô tô nghỉ 15 phút và cả thời gian ca đi cả về là 2h30'

   \(\frac{x}{30}+\frac{x}{40}+\frac{1}{4}=\frac{5}{2}\)

\(\Leftrightarrow\)\(\frac{x}{30}+\frac{x}{40}=\frac{5}{2}-\frac{1}{4}\)

\(\Leftrightarrow\)\(\frac{x}{30}+\frac{x}{40}=\frac{9}{4}\)

\(\Leftrightarrow\)\(\frac{4x}{120}+\frac{3x}{120}=\frac{270}{120}\)

\(\Leftrightarrow\)\(7x=270\)\(\)

\(\Leftrightarrow\)\(x=30\)

Vậy quãng đường AB là 30 km.

2) \(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{2011}{3}\)

\(\Leftrightarrow\)\(\frac{x-3}{2011}-1+\frac{x-2}{2012}-1=\)\(\frac{x-2012}{2}-1+\frac{x-2011}{3}-1\)

\(\Leftrightarrow\)\(\frac{x-3}{2011}-\frac{2011}{2011}+\frac{x-2}{2012}-\frac{2012}{2012}=\)\(\frac{x-2012}{2}-\frac{2}{2}+\frac{x-2011}{3}-\frac{3}{3}\)

\(\Leftrightarrow\)\(\frac{x-2014}{2011}+\frac{x-2014}{2012}=\)\(\frac{x-2014}{2}+\frac{x-2014}{3}\)

\(\Leftrightarrow\left(x-2014\right).\)\(\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)\)\(=0\)

Vì  \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}>0\)

\(\Leftrightarrow\)\(x-2014=0\)

\(\Leftrightarrow\)\(x=2014\)

Vậy phương trình có nghiệm là : x = 2014

Hi everybody, I represent for my group to report about the party to welcome the new year. The main content of the party is that we will organize a party to welcome the new year at the restaurant at 7am and 7pm. January, 2013. We invite the people to attend our party. All friends in the class will be invited. At 8am, we will organize the contest of packing the rice cake. The winner will be received a special reward – one of three red envelops. In these three envelops, there will be an envelop with 20 dollars, 1 with 10 dollar, and 5 dollar. The reward will be rewarded in the second party at 7pm in the same day. Especially, everybody attending the party will be received a red envelop as the first year reward. Beginning the second phase party is “Happy New Year” song and a clapping wave. Then, there will be some special foods and drinks prepared by me such as beefsteak, roll grill, soft drink, fruits, cake, fried potatoes, candy and melon seed, so on and some special foods as the rice cake packed in the morning. And to join in the fun for the party, we will invite a international jugglery to perform, his name is Luu Khiem. And the comedian that we refer his name, anyone must laugh, it is Hoai Linh comedian. Finally, it will be the rewarding to the rice cake packer. Please inform me 1 week before the party if you can attend.