a) Ta có: \(P=\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{3\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}+1}\right)\)

\(=\dfrac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{3\sqrt{a}\left(\sqrt{a}+1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)

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

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

\(=\dfrac{2\sqrt{a}+2a+2\sqrt{a}+2}{\sqrt{a}}\)

\(=\dfrac{2\left(a+2\sqrt{a}+1\right)}{\sqrt{a}}\)

\(=\dfrac{2\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)

b) Ta có: \(P-6=\dfrac{2\left(\sqrt{a}+1\right)^2-6\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2a+4\sqrt{a}+2-6\sqrt{a}}{\sqrt{a}}\)

\(=\dfrac{2\left(a-\sqrt{a}+1\right)}{\sqrt{a}}>0\forall a\) thỏa mãn ĐKXĐ

hay P>6

Part 1

1 into

2 which

3 happily

4 aspect

5 if

6 compulsory

Part 2

1c 2f 3b 4g 5a 6e

Part 3

1 opened

2 to be finished

3 repairing

4 driving

5 harmful

6 modernize

7 environmentalists

8 effectively

Part 4

1 disappointing -> disappointed

2 come -> came

Part 5

1 succeed although he tried 

2 my father could speak

3 seen such an interesting 

4 being made to protect

Part 6

1 forest

2 climate

3 in

4 fortunatelt

Part 7

1T 2F 3F 4F

Đặt \(log_2x=t\Rightarrow t\ge4\)

Phương trình trở thành: \(\sqrt{t^2-2t-3}=m\left(t-3\right)\)

\(\Leftrightarrow\sqrt{\left(t+1\right)\left(t-3\right)}=m\left(t-3\right)\)

\(\Leftrightarrow\sqrt{t+1}=m\sqrt{t-3}\)

\(\Leftrightarrow m=\sqrt{\dfrac{t+1}{t-3}}\)

Hàm \(f\left(t\right)=\sqrt{\dfrac{t+1}{t-3}}\) nghịch biến khi \(t\ge4\)

\(\lim\limits_{t\rightarrow+\infty}\sqrt{\dfrac{t+1}{t-3}}=1\) ; \(f\left(4\right)=\sqrt{5}\)

\(\Rightarrow1< f\left(t\right)\le\sqrt{5}\Rightarrow1< m\le\sqrt{5}\)

Đáp án D

a: Xét (O) có 

ΔABC nội tiếp đường tròn

AB là đường kính

Do đó: ΔABC vuông tại C