Question

 

cho em xin bài 3 câu 3 rở đi ạ

Answer 1

3:

1: Khi x=6+2 căn 5 thì \(A=\dfrac{\sqrt{5}+1+3}{\sqrt{5}+1-2}=\dfrac{4+\sqrt{5}}{\sqrt{5}-1}\)

2: \(=\dfrac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4x}{x-4}=\dfrac{4x+8\sqrt{x}}{x-4}=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\)

3: \(P=B:A\)

\(=\dfrac{4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+3}=\dfrac{4\sqrt{x}}{\sqrt{x}+3}=4-\dfrac{12}{\sqrt{x}+3}\)

Để P nguyên thì \(\sqrt{x}+3\in\left\{3;4;6;12\right\}\)

=>\(x\in\left\{0;1;9;81\right\}\)