a) Bạn dư sức làm.
b) \(A=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\cdot\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\cdot\left(\sqrt{x}+1\right)\cdot\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\sqrt{x}\cdot\left(\sqrt{x}+1\right)+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=x+\sqrt{x}+1-\dfrac{2x+\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}+x+\sqrt{x}-\left(2x+\sqrt{x}\right)}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}+x+\sqrt{x}-2x-\sqrt{x}}{\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-x}{\sqrt{x}}\)
\(=\dfrac{\left(x\sqrt{x}-x\right)\sqrt{x}}{x}\)
\(=\dfrac{x\cdot\left(\sqrt{x}-1\right)\sqrt{x}}{x}\)
\(=\left(\sqrt{x}-1\right)\sqrt{x}\)
\(=x-\sqrt{x}\)
