Bạn thử làm xem nếu câu hỏi nào không biết thì mới hỏi nha, chứ đăng như này làm hơi dài á với lại mình thấy bạn hay đăng mấy dạng toán như này, bạn chỉ cần làm tương tự như nhưng bài trước là được :<
a. \(P=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\right)\)

\(=\dfrac{x+\sqrt{x}-x-2}{\sqrt{x}+1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x-4}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)

b. Để \(P< \dfrac{1}{2}\) thì \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}< \dfrac{1}{2}\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-4}{2\left(\sqrt{x}+2\right)}< 0\Leftrightarrow\dfrac{\sqrt{x}-4}{2\left(\sqrt{x}+2\right)}>0\)

ta có \(\left\{{}\begin{matrix}\sqrt{x}-4< 0\\2\left(\sqrt{x}+2\right)\ge4>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x}>4\Rightarrow x< 16\)

c. \(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{1-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

Ta có với mọi \(x\ge0\) thì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\Rightarrow P=1-\dfrac{3}{\sqrt{x}+2}\le-\dfrac{1}{2}\) 

Dấu "=" xảy ra khi \(x=0\)

Vậy \(P_{min}=-\dfrac{1}{2}\) khi \(x=0\)

d. Ta có \(Q=P\left(2\sqrt{x}+x\right)=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+x\right)}{\sqrt{x}+2}=\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}\left(2+\sqrt{x}\right)}{\sqrt{x}+2}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\forall x\)

Dấu "=" xảy ra khi \(x=\dfrac{1}{4}\)

Vậy \(Q_{min}=-\dfrac{1}{4}\) khi \(x=\dfrac{1}{4}\)

ò thanks :>

1. marvelous

2. developments

3. researcher 

4. encourage 

5. curiosity 

6. optimisticc 

7. weightless 

8. incredible

The girl, who sits next to me is intelligent

Mình làm hết r nha, mỏi tay ghê :<

Có bạn làm 1234 r nên mình làm từ bài 5 trở đi nha =))
5. \(\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}=\dfrac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

6. \(\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\dfrac{5-2\sqrt{5}}{2\sqrt{5}-4}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{2\left(\sqrt{5}-2\right)}=\sqrt{5}+\dfrac{\sqrt{5}}{2}=\dfrac{3\sqrt{5}}{2}\)

7. \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\dfrac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\dfrac{2\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)

8. \(\dfrac{3\sqrt{2}-6}{\sqrt{2}-1}+\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}=\dfrac{3\sqrt{2}\left(1-\sqrt{2}\right)}{\sqrt{2}-1}+\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{\sqrt{2}-3}=-3\sqrt{2}-2\sqrt{2}=-5\sqrt{2}\)

9. \(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)=\dfrac{\sqrt{3}\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)=\sqrt{3}+2+\sqrt{2}-\left(2+\sqrt{3}\right)=\sqrt{2}\)

10. \(\dfrac{\sqrt{160}-\sqrt{80}}{\sqrt{8}-\sqrt{2}}-\dfrac{\sqrt{40}-\sqrt{15}}{2\sqrt{2}-\sqrt{3}}=\dfrac{\sqrt{2}\left(4\sqrt{5}-2\sqrt{10}\right)}{\sqrt{2}}-\dfrac{\sqrt{5}\left(2\sqrt{2}-\sqrt{3}\right)}{2\sqrt{2}-\sqrt{3}}=4\sqrt{5}-2\sqrt{10}-\sqrt{5}=3\sqrt{5}-2\sqrt{10}\)

11. \(\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)=-\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)=-1+5=4\)

12. \(\left(\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)=-\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)=-5+4=-1\)

\(K=\left\{0;1;2;3;4;5;6\right\}\)

Do you know the man who is talking to Tom?