1, heroic

2, cruelty

3, candidacy

4, disconnect

5, monotonous

6, unrealistic

7, occurrence

8, publication

5) 

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

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

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

4)

\(B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

3) \(B=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)

2) 

\(B=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+7\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\dfrac{2x-6\sqrt{x}+3+x+4\sqrt{x}-7\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\dfrac{3x-9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+3}\)

1)

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

\(B=\dfrac{2\sqrt{x}-2\sqrt{x}+6}{x-9}=\dfrac{6}{x-9}\)

Bạn ơi con này ms đúng ạ. Lúc nãy mk lm sai

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