Question

Cho P \(=\dfrac{2}{x+\sqrt{x}+2}\) với x ≥ 0, giá trị của x để P = \(\dfrac{1}{2}\)

Answer 1

Ta có:

\(P=\dfrac{1}{2}\) khi:

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

\(\Rightarrow2\cdot2=1\cdot\left(x+\sqrt{x}+2\right)\)

\(\Leftrightarrow4=x+\sqrt{x}+2\)

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

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

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

\(\Leftrightarrow\left(x+2\sqrt{x}\right)-\left(\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\)

Mà: \(\sqrt{x}\ge0\forall x\Rightarrow\sqrt{x}+2\ge2\forall x\)

\(\Leftrightarrow\sqrt{x}-1=0\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1^2\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy: \(x=1\)

Answer 2

P=1/2

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

=>\(x+\sqrt{x}+2=4\)

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

=>\(\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)=0\)

=>\(\sqrt{x}-1=0\)

=>x=1