1.Her uncle is always visited on Sunday by her.

2.Her aunt wasn't went to visit last Sunday by her.

3.I was called to join him last month by him.

4.Rice in the paddy field is being planted at present by my parents.

5.Japanese food was already tried before by her.

6.Books are being read at present by Nam

để mik nghĩ đã,nhưng có lẽ ko có

B.Nhà Hán

Bổ đề :\(\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge9\)

Áp dụng bất đẳng thức Cô-si ta có:

 \(x+y+z\ge3\sqrt[3]{xyz};\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge3\sqrt[3]{\dfrac{1}{x}.\dfrac{1}{y}.\dfrac{1}{z}}\)

\(\Rightarrow\left(x+y+z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{\dfrac{1}{x}\dfrac{1}{y}\dfrac{1}{z}}=9\) 

Dấu "=" xảy ra ⇔ x=y=z

Ta có:\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{9}.\dfrac{9}{a+3b+2c}\le\dfrac{ab}{9}.\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)\)

Tương tự ta có:\(\dfrac{bc}{b+3c+2a}\le\dfrac{bc}{9}\left(\dfrac{1}{b+a}+\dfrac{1}{c+a}+\dfrac{1}{2c}\right)\)

                         \(\dfrac{ca}{c+3a+2b}\le\dfrac{ca}{9}.\left(\dfrac{1}{c+b}+\dfrac{1}{a+b}+\dfrac{1}{2a}\right)\)

Cộng vế với vế ta có:

\(A\le\dfrac{1}{9}.\left(\dfrac{ab+bc}{a+c}+\dfrac{cb+ac}{a+b}+\dfrac{ca+ab}{b+c}+\dfrac{a+b+c}{2}\right)\)

\(=\dfrac{1}{9}.\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}.\left(6+\dfrac{6}{3}\right)=1\)

Dấu "=" xảy ra ⇔ a=b=c=2

Vậy Max A=1⇔ a=b=c=2

Ta có:a^{4 m}b^{4 m}-(a ^mb ^m+1)(a^{2 m}b^{2 m}+1)(a ^mb ^m-1)

   =  a^{4 m}b^{4 m}-(a ^2mb ^2m-1)(a^{2 m}b^{2 m}+1)

   =  a^{4 m}b^{4 m}-(a ^4mb ^4m-1)

   = 1

Cách đặt mặt lớn nhất của hộp phấn tiếp xúc với bàn thì tạo ra ma sát lớn hơn cách còn lại

nhân hóa

1.It took them a lot of time to turn off this machine

2.Unless I leave now,I'll miss the train.

3.Although it is raining,we still go out.

4.Jane is not keen to watch sports on TV.

5.I spent the whole day making the plan.

cop nhưng vẫn tick