a, \(A=\dfrac{\sqrt{x}-1}{x^2-x}:\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)=\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}\pm1\right)}:\left(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{1}{x\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}=1\)
b, Cho A = 1 rồi còn gì, hay đề lỗi bạn ?
\(x=4+2\sqrt{3}=\sqrt{3}^2+2\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
xem là bài mình làm có sai đâu ko nhé nếu rút gọn ra kq khác thì thay bên trên vào nhé
a) Ta có: \(A=\dfrac{\sqrt{x}-1}{x^2-x}:\left(\dfrac{1}{\sqrt{x}}-\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\sqrt{x}-1}{x\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1}\)
=1