\(B=\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)
\(B=\dfrac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}\)
\(B=\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
\(B=x-1\)
\(B=A+1\Leftrightarrow\sqrt{x}-1+1=x-1\)
\(\Leftrightarrow x-\sqrt{x}-1=0\)
\(\Leftrightarrow x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}-1=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}\right)\left(\sqrt{x}-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\left(\sqrt{5}+1\right)^2}{4}\\x=\dfrac{\left(1-\sqrt{5}\right)^2}{4}\end{matrix}\right.\)
câu A sửa lại đề 1 chút
\(A=\dfrac{x-3\sqrt{x}+2}{\sqrt{x}-2}\)
\(A=\dfrac{x-2\sqrt{x}-\sqrt{x}+2}{\sqrt{x}-2}\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(A=\sqrt{x}-1\)
có \(x=4-2\sqrt{3}\)
\(\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
\(\Leftrightarrow\sqrt{x}=\sqrt{3}-1\)
khi đó \(A=\sqrt{x}-1\Leftrightarrow A=\sqrt{3}-1-1=\sqrt{3}-2\)
em cảm mơn nhìu nhìu 
