Question

giúp mình bài 4 với ạ mình cảm ơn

Answer 1

b1 x khác 4

Answer 2

4:

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

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

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

b: |A|>A

=>A<0

=>\(\dfrac{-3}{\sqrt{x}-2}< 0\)

=>căn x-2>0

=>x>4

c: \(B=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}+3}\cdot\dfrac{-3}{\left(\sqrt{x}-2\right)}=\dfrac{-6}{\sqrt{x}+3}\)

Để B là số nguyên thì \(\sqrt{x}+3\inƯ\left(-6\right)\)

=>\(\sqrt{x}+3\in\left\{3;6\right\}\)

=>\(x\in\left\{0;9\right\}\)

mà x<>9

nên x=0