Toán Lớp 9: $B={\huge{(}}\dfrac{1}{\sqrt{1}}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}{\huge{)}}:\dfrac{\sqrt{x}}{x+\sqrt{x}}\ \ (x>0)\\=\dfrac{x+1}{\sqrt{x}}$
Tìm $x$ để $B=2$
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Toán Lớp 9: $B={\huge{(}}\dfrac{1}{\sqrt{1}}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}{\huge{)}}:\dfrac{\sqrt{x}}{x+\sqrt{x}}\ \ (x>0)\\=\dfrac{x+1}{\sqrt{x}}$
Tìm $x$ để $B=2$
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