Question

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

tìm x để bt =4

Answer 1

\(A=\dfrac{x+2}{\sqrt{x}}.\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)                  \(ĐK:x\ge0\)

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

\(A=4\Leftrightarrow\dfrac{2\left(x+2\right)}{\sqrt{x}+1}=4\Leftrightarrow2\left(x+2\right)=4\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow2x+4=4\sqrt{x}+4\)

\(\Leftrightarrow\sqrt{x}=2\)

\(\Leftrightarrow x=4\)