Question

tìm \(x\\ \):A= \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

biết A=5

Answer 1

Ta có:

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

Để \(A=5\)

\(\Leftrightarrow1+\dfrac{2}{\sqrt{x}-1}=5\Leftrightarrow\dfrac{2}{\sqrt{x}-1}=4\Leftrightarrow\sqrt{x}-1=\dfrac{1}{2}\Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\)

Vậy \(x=\dfrac{9}{4}\) thì \(A=5\).

Answer 2

\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=5\)

\(\Rightarrow\sqrt{x}+1=\left(\sqrt{x}-1\right).5\)

\(\Rightarrow\sqrt{x}+1=5.\sqrt{x}-5\)

\(\Rightarrow-4\sqrt{x}=-6\)

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

\(\Rightarrow x=\dfrac{9}{4}\)