Question

Answer 1

Giúp mình ý d bài 2 với ạ

 

Answer 2

a) \(P=\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\) \(\left(đk:x>1,x\ne0\right)\)

\(P=\dfrac{2}{\sqrt{x}-1}\)

b) \(x=\dfrac{4}{9}\left(tmđk\right)\)

\(\Rightarrow\sqrt{x}=\dfrac{2}{3}\)

\(P=\dfrac{2}{\dfrac{2}{3}-1}=-6\)

\(x=3+\sqrt{8}\left(tmđk\right)\)

Answer 3

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

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

\(=\dfrac{2x+2}{x-1}\)