1: \(P=\left(\dfrac{\sqrt{x-2}\left(3-\sqrt{x-2}\right)}{9-x+2}+\dfrac{x+7}{11-x}\right):\left(\dfrac{3\sqrt{x-2}+1-\sqrt{x-2}+3}{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}\right)\)

\(=\dfrac{3\sqrt{x-2}-x+2+x+7}{11-x}:\dfrac{2\sqrt{x-2}+4}{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}\)

\(=\dfrac{3\sqrt{x-2}+9}{11-x}\cdot\dfrac{\sqrt{x-2}\left(\sqrt{x-2}-3\right)}{2\sqrt{x-2}+4}\)

\(=\dfrac{-3\left(\sqrt{x-2}+3\right)}{2\left(\sqrt{x-2}+2\right)}\cdot\dfrac{\sqrt{x-2}}{\sqrt{x-2}+3}\)

\(=\dfrac{-3\sqrt{x-2}}{2\sqrt{x-2}+4}\)

2: Đặt căn x-2=a(a>=0)

=>P=-3a/(2a+4)

P nguyên

=>-3a chia hết cho 2a+4

=>-6a chia hết cho 2a+4

=>-6a-12+12 chia hết cho 2a+4

=>2a+4 thuộc {1;-1;2;-2;3;-3;4;-4;6;-6;12;-12}

=>a thuộc {1;4}

=>x-2=1 hoặc x-2=16

=>x=3 hoặc x=18

8 How much do these apples cost?

9 THis is a blue car

11 Are there 40 classrooms in Phong's school?

13 How wide if the Great wall

1 send - will receive 

2 do - will improve

3 find - will give

4 will go - has

5 will go- gets

6 doesn't phone - will leave

7 don't study - won't pass

8 rains - won't have to

9 won't be able - watch

10 can't move - isn't

11 study - will pass

Lỗi r, xem lại nhé

1.

a, \(\left(C\right)x^2+y^2-6x-2y+6=0\)

\(\Leftrightarrow\left(C\right)\left(x-3\right)^2+\left(y-1\right)^2=4\)

\(\Rightarrow\) Tâm \(I=\left(3;1\right)\), bán kính \(R=2\)

b, Tiếp tuyến đi qua A có dạng: \(\left(\Delta\right)ax+by-5a-7b=0\left(a^2+b^2\ne0\right)\)

Ta có: \(d\left(I;\Delta\right)=\dfrac{\left|3a+b-5a-7b\right|}{\sqrt{a^2+b^2}}=2\)

\(\Leftrightarrow\left|a+3b\right|=\sqrt{a^2+b^2}\)

\(\Leftrightarrow6ab+8b^2=0\)

\(\Leftrightarrow2b\left(3a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=0\\3a+4b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\Delta_1:x=5\\\Delta_2:4x-3y+1=0\end{matrix}\right.\)

TH1: \(\Delta_1:x=5\)

Tiếp điểm có tọa độ là nghiệm hệ: \(\left\{{}\begin{matrix}x=5\\x^2+y^2-6x-2y+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y^2-2y+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=1\end{matrix}\right.\Rightarrow\left(5;1\right)\)

TH2: \(\Delta_2:4x-3y+1=0\)

Tiếp điểm có tọa độ là nghiệm hệ: \(\left\{{}\begin{matrix}4x-3y+1=0\\x^2+y^2-6x-2y+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{5}\\y=\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left(\dfrac{7}{5};\dfrac{11}{5}\right)\)

Kết luận: Phương trình tiếp tuyến: \(\left\{{}\begin{matrix}\Delta_1:x=5\\\Delta_2:4x-3y+1=0\end{matrix}\right.\)

Tọa độ tiếp điểm: \(\left\{{}\begin{matrix}\left(5;1\right)\\\left(\dfrac{7}{5};\dfrac{11}{5}\right)\end{matrix}\right.\)

a computer is used to do that job nowadays

He will be seen off at the airport by all his friends 

Beer used to be drunk for breakfast in England years ago

Tea can not made with cold water.

the floor was being cleaned when i arrived

Should Julia be helped with the sewing ?