Question

Tìm x để \(A=\dfrac{1}{2}\) biết \(A=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)

Answer 1

ĐKXĐ: \(x>0\)

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

\(\Leftrightarrow2\cdot\left(\sqrt{x}-2\right)=\sqrt{x}\cdot1\)

\(\Leftrightarrow2\sqrt{x}-4=\sqrt{x}\)

\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=4\)

\(\Leftrightarrow\sqrt{x}=4\)

\(\Leftrightarrow x=16\left(tm\right)\)

Vậy \(A=\dfrac12\) khi \(x=16\).

Answer 2

ĐKXĐ: x>0

\(A=\dfrac{1}{2}\)

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

=>\(2\left(\sqrt{x}-2\right)=\sqrt{x}\)

=>\(2\sqrt{x}-4-\sqrt{x}=0\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

Answer 3

A=12�=12 khi $x=16$.