Câu hỏi của Bachifuzuha

a: \(B=\dfrac{x+1-2\sqrt{x}}{x-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)\(C=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)b: B*C=1/3=>\(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{3}\)=>\(3\left(x-2\sqrt{x}+1\right)=x+\sqrt{x}\)=>\(2x-7\sqrt{x}+3=0\)=>\(\left(\sqrt{x}-3\right)\left(2\sqrt{x}-1\right)=0\)=>x=9 hoặc x=1/2 Câu trả lời: a: \(B=\dfrac{x+1-2\sqrt{x}}{x-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)\(C=\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)b: B*C=1/3=>\(\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}+1\right)}=\dfrac{1}{3}\)=>\(3\left(x-2\sqrt{x}+1\right)=x+\sqrt{x}\)=>\(2x-7\sqrt{x}+3=0\)=>\(\left(\sqrt{x}-3\right)\left(2\sqrt{x}-1\right)=0\)=>x=9 hoặc x=1/2