a) \(A=\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(A=\left[\dfrac{2x+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\left[\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right]\)
\(A=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(x-\sqrt{x}+1-\sqrt{x}\right)\)
\(A=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(x-2\sqrt{x}+1\right)\)
\(A=\dfrac{1}{\sqrt{x}-1}\cdot\left(\sqrt{x}-1\right)^2\)
\(A=\sqrt{x}-1\)
b) \(A=3\) khi:
\(\sqrt{x}-1=3\)
\(\Rightarrow\sqrt{x}=3+1\)
\(\Rightarrow\sqrt{x}=4\)
\(\Rightarrow x=16\left(tm\right)\)
