\(Q=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\left(dkxdx\ge0\right)\)
THay `Q = 1/2` có :
\(\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{1}{2}\)
`=>`\(2\cdot\left(2-5\sqrt{x}\right)=\sqrt{x}+3\)
`=>`\(4-10\sqrt{x}=\sqrt{x}+3\)
`=> `\(-10\sqrt{x}-\sqrt{x}=-4+3\)
`=>`\(-11\sqrt{x}=-1\)
`=>`\(\sqrt{x}=\dfrac{1}{11}\)
`=> \(x=\dfrac{1}{121}\)(thỏa mãn dkxd)
Vậy `x=1/121`