ĐKXĐ: \(x\ge0\)

Ta có \(P=\dfrac{\sqrt{x}+7}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2+5}{\sqrt{x}+2}=1+\dfrac{5}{\sqrt{x}+2}\) 

Để P nguyên thì \(5⋮\left(\sqrt{x}+2\right)\Rightarrow\left(\sqrt{x}+2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+2=1\\\sqrt{x}+2=-1\\\sqrt{x}+2=5\\\sqrt{x}+2=-5\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=-1\\\sqrt{x}=-3\\\sqrt{x}=3\\\sqrt{x}=-7\end{matrix}\right.\) \(\Rightarrow x=9\) ( thỏa mãn đk )

Vậy x thỏa mãn ycbt là x=9

C

Rewrite the following sentences as directed. (10)

1.     The children couldn’t go swimming because the sea was very rough.

→ The sea was very rough so the children couldn't go swimming

2.     She started working as a secretary last Christmas.               

→ She has worked as a secretary since last Chistmas

3.     It is necessary to do all these exercises before the exam.    

→ You need to do all these exercises before the exam 

4.     How long is it since they bought that house?                       

→ When did they buy that house ?

5.      Although his both legs were broken in the crash, he managed to get out of the car before it exploded.

→ Despite his both broken legs in the crash, he managed to get out of the car before it exploded

6.      “ How many survivors are there?”, asked the journalist.

→ The journalist wanted to know how many survivors there were

7.      It was such rotten meat that it had to be thrown away.

→ The meat was so rotten that it had to be thrown away

8.      You can’t visit the USA unless you have a visa.

→ If you have a visa, you can visit the USA

9.      She knows a lot more about it than I do.

→ I don’t know as much about it as she does

10.    My French friend finds driving on the left difficult.

→ My French friend isn’t used to driving on the left

b. \(\sqrt{3x}+5\sqrt{27x}-16=\sqrt{432x}\) (*)

ĐKXĐ: \(x\ge0\)

(*) \(\Leftrightarrow\sqrt{3x}+5\sqrt{27x}-\sqrt{432x}=16\)

\(\Leftrightarrow\sqrt{3x}\left(1+5\sqrt{9}-\sqrt{144}\right)=16\)

\(\Leftrightarrow\sqrt{3x}\left(5\sqrt{9}+1-12\right)=16\)

\(\Leftrightarrow4\sqrt{3x}=16\)

\(\Leftrightarrow\sqrt{3x}=4\)

\(\Leftrightarrow3x=16\)

\(\Leftrightarrow x=\dfrac{16}{3}\) ( thỏa mãn đk )

Vậy \(s=\left\{\dfrac{16}{3}\right\}\)

Giúp câu c với ạ

Ta có \(A=\dfrac{4x-3}{x+2}=\dfrac{4x+8-11}{x+2}=4-\dfrac{11}{x+2}\)

Để \(A\) nguyên thì \(11⋮\left(x+2\right)\Rightarrow\left(x+2\right)\inƯ\left(11\right)=\left\{1;-1;11;-11\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+2=1\\x+2=-1\\x+2=11\\x+2=-11\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=9\\x=-13\end{matrix}\right.\)

Vậy tất cả các x thỏa ycbt là x=-1;x=-3;x=9 hoặc x=-13

11. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)  

12. \(\sqrt{3+2\sqrt{2}}+\sqrt{5-2\sqrt{6}}\)

\(=\sqrt{\left(1+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=1+\sqrt{2}+\sqrt{3}-\sqrt{2}=1+\sqrt{3}\)

13. \(\sqrt{9-4\sqrt{5}}+\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\sqrt{5}-2+\sqrt{5}+1=2\sqrt{5}-1\)

14. \(\sqrt{11+6\sqrt{2}}+\sqrt{6+4\sqrt{2}}-2\sqrt{2}\)

\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}-2\sqrt{2}\)

\(=3+\sqrt{2}+2+\sqrt{2}-2\sqrt{2}=5\)

15. \(\sqrt{15-6\sqrt{6}}+\sqrt{10+4\sqrt{6}}-\sqrt{7+2\sqrt{6}}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2+\sqrt{6}\right)^2}-\sqrt{\left(1+\sqrt{6}\right)^2}\)

\(=3-\sqrt{6}+2+\sqrt{6}-1-\sqrt{6}=4-\sqrt{6}\)

16. \(\sqrt{31-8\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

\(=4-\sqrt{15}+\sqrt{15}-3=1\)

1. \(\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

2. \(\sqrt{7-4\sqrt{3}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

3. \(\sqrt{5+2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}\)

4. \(\sqrt{16+6\sqrt{7}}=\sqrt{\left(3+\sqrt{7}\right)^2}=3+\sqrt{7}\)

5. \(\sqrt{12+2\sqrt{35}}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}=\sqrt{7}+\sqrt{5}\)

6. \(\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

7. \(\sqrt{18-2\sqrt{65}}=\sqrt{\left(\sqrt{13}-\sqrt{5}\right)^2}=\sqrt{13}-\sqrt{5}\)

8. \(\sqrt{27+10\sqrt{2}}=\sqrt{\left(5+\sqrt{2}\right)^2}=5+\sqrt{2}\)

9. \(\sqrt{14+6\sqrt{5}}=\sqrt{\left(3+\sqrt{5}\right)^2}=3+\sqrt{5}\)

10. \(\sqrt{3+2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{\left(1+\sqrt{2}\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}=1+\sqrt{2}-2+\sqrt{2}=2\sqrt{2}-1\)

\(\left(2x-1\right)^5=x^5\)

\(\Leftrightarrow2x-1=x\)

\(\Leftrightarrow x=1\)