1. Đ

2. S

3.Đ

bạn ơi còn 4,5,6 nx giúp mik vs

b: Xét ΔAHB vuông tại H có HE là đường cao

nên \(AE\cdot AB=AH^2\left(1\right)\)

Xét ΔAHC vuông tại H có HF là đường cao

nên \(AF\cdot AC=AH^2\left(2\right)\)

Từ (1) và (2) suy ra \(AE\cdot AB=AF\cdot AC\)

a, đkxđ : a >= 0 ; a khác 9 

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

\(=\dfrac{2a-6\sqrt{a}+a+4\sqrt{a}+3-3-7\sqrt{a}}{a-9}=\dfrac{3a-9\sqrt{a}}{a-9}=\dfrac{3\sqrt{a}}{\sqrt{a}+3}\)

b, \(\dfrac{3\sqrt{a}}{\sqrt{a}+3}=\dfrac{3\left(\sqrt{a}+3\right)-9}{\sqrt{a}+3}=3-\dfrac{9}{\sqrt{a}+3}\Rightarrow\sqrt{a}+3\inƯ\left(9\right)=\left\{1;3;9\right\}\)

c, Ta có : \(3=\dfrac{3\left(\sqrt{a}+3\right)}{\sqrt{a}+3}\)mà \(3\left(\sqrt{a}+3\right)>3\sqrt{a}\)

Vậy P < 3 

d, Ta có : \(\dfrac{3\sqrt{a}}{\sqrt{a}+3}.\dfrac{\sqrt{a}-1}{\sqrt{a}}< 1\Leftrightarrow\dfrac{3\left(\sqrt{a}-1\right)}{\sqrt{a}+3}-1< 0\)

\(\Leftrightarrow\dfrac{3\sqrt{a}-3-\sqrt{a}-6}{\sqrt{a}+3}< 0\Rightarrow2\sqrt{a}-9< 0\)vì \(\sqrt{a}+3>0\)

\(\Leftrightarrow\sqrt{a}< \dfrac{9}{2}\Leftrightarrow a< \dfrac{81}{4}\)

Kết hợp đk vậy \(0\le a< \dfrac{81}{4}\)

\(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)

\(=\dfrac{3\sqrt{x}-9+x+9}{x-9}:\dfrac{2\sqrt{x}-5-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

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

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