`#tnvt`
`S=\frac{1+\sqrt{x}}{\sqrt{x}}(x>0)`
`d)`
`2S=2\sqrt{x}+5`
`=>2.\frac{1+\sqrt{x}}{\sqrt{x}}=2\sqrt{x}+5`
`<=>\frac{2+2\sqrt{x}}{\sqrt{x}}=2\sqrt{x}+5`
`<=>2+2\sqrt{x}=(2\sqrt{x}+5).\sqrt{x}`
`<=>2+2\sqrt{x}=2x+5\sqrt{x}`
`<=>2x+5\sqrt{x}-2\sqrt{x}-2=0`
`<=>2x+3\sqrt{x}-2=0`
`<=>2x+4\sqrt{x}-\sqrt{x}-2=0`
`<=>2\sqrt{x}(\sqrt{x}+2)-(\sqrt{x}+2)=0`
`<=>(2\sqrt{x}-1)(\sqrt{x}+2)=0`
`<=>[(2\sqrt{x}-1=0),(\sqrt{x}+2=0):}`
`<=>[(2\sqrt{x}=1),(\sqrt{x}=-2(\text{Vô lý})):}`
`<=>\sqrt{x}=1/2`
`<=>x=1/4(\text{Thỏa mãn})`
Vậy `x=1/4` thì `2S=2\sqrt{x}+5`
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`\text{Bạn tus bảo chỉ làm câu d, ở phần bl nhé.}`
