Question

\(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=3\)

Giải pt với x>0

Giúp với mai thi

Answer 1

\(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=3\)

\(\Leftrightarrow\dfrac{2\sqrt{x}\sqrt{x}}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}=\dfrac{3.2\sqrt{x}}{2\sqrt{x}}\)

\(\Leftrightarrow\dfrac{2x}{2\sqrt{x}}-\dfrac{6\sqrt{x}}{2\sqrt{x}}+\dfrac{1}{2\sqrt{x}}=0\)

\(\Leftrightarrow2x-6\sqrt{x}+1=0\)

\(\Leftrightarrow...\)

Answer 2

\(\Rightarrow2x+1=6\sqrt{x}\)

\(\Rightarrow2x-6\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\dfrac{3\pm\sqrt{7}}{2}\)

\(\Rightarrow x=\left(\dfrac{3\pm\sqrt{7}}{2}\right)^2=\dfrac{8\pm3\sqrt{7}}{2}\)