1. A : Someone has locked the door

=> P : The door has been locked 

2. P : Ann was invited to the party yesterday

=> A : Someone invited Ann to the party yesteday

3. A : She made a mistake in the examination

=> P : A mistake was made in the examination by her

4. P : Football is played all over the world

=> A : Everyone plays football all over the world

5. A : Our country exports rice and sugar

=> P : Rice and sugar are exported by our country

"Does Nga play volleyball?" => No, she doesn't

I school is bigger than her school

=> My school is bigger than her school

Ta có:

\(x^2+2x-7=x\left(x+2\right)-7\)

Mà \(x\left(x+2\right)\)chia hết cho \(x+2\)

\(\Rightarrow7\)chia hết cho \(x+2\)

hay \(x+2\)\(\in\)Ư(7)

Ta có bảng sau:

We can listening to music on the radio

=> Sửa lại : We can listen to music on the radio

a/ Vì \(x\ge3>-\frac{2}{3}\) nên giá trị biểu thức là : 

\(x+\frac{2}{3}+x-3=2x-\frac{7}{3}\)

b/ Vì \(x>2>\frac{4}{3}>-\frac{2}{5}\) nên giá trị biểu thức là : 

\(-\left(x+\frac{2}{5}\right)+\left(x-\frac{4}{3}\right)=-\frac{4}{3}-\frac{2}{5}=-\frac{26}{15}\)

Đặt \(y=\left(x-3\right)\sqrt{\frac{x+1}{x-3}}\)

Suy ra pt trở thành \(y^2+y+3=0\)

Mà : \(y^2+y+3=\left(y^2+y+\frac{1}{4}\right)+\frac{11}{4}=\left(y+\frac{1}{2}\right)^2+\frac{11}{4}>0\)

Do đó pt trên vô nghiệm.

Ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+zx=xyz\)

\(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\ge\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)

Bình phương vế trái : 

\(\left(\sqrt{x+yz}+\sqrt{y+zx}+\sqrt{z+xy}\right)^2\)

\(=\left(x+y+z+xy+yz+zx\right)+2\left(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\right)\)Bình phương vế phải : 

\(\left(\sqrt{xyz}+\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=\left(xyz+x+y+z\right)+2\left(x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)

Suy ra cần phải chứng minh : \(\sqrt{x+yz}.\sqrt{y+zx}+\sqrt{y+zx}.\sqrt{z+xy}+\sqrt{z+xy}.\sqrt{x+yz}\ge x\sqrt{yz}+y\sqrt{xz}+z\sqrt{xy}+\sqrt{x}+\sqrt{y}+\sqrt{z}\)(*)

Thật vậy, theo bđt Bunhiacopxki ta có : \(\sqrt{x+yz}.\sqrt{y+zx}\ge\sqrt{xy}+z\sqrt{xy}\)

\(\sqrt{y+zx}.\sqrt{z+xy}\ge\sqrt{yz}+x\sqrt{yz}\)

\(\sqrt{z+xy}.\sqrt{x+yz}\ge\sqrt{xz}+y\sqrt{xz}\)

Cộng các bđt trên theo vế ta chứng minh được (*) đúng.

Vậy bđt ban đầu được chứng minh.

a/ \(M_{Cl_2}=35,5.2=71\)

b/ \(M_{H_2SO_4}=1.2+32+16.4=98\)

c/ \(M_{KMnO_4}=39+55+16\cdot4=158\)

Cacbon dioxit (CO₂) gồm hai một nguyên tử C liên kết với hai nguyên tử O