a:
Sửa đề: Tính giá trị của A khi x=25
Khi x=25 thì \(A=\dfrac{\sqrt{25}+10}{\sqrt{25}}=\dfrac{5+10}{5}=\dfrac{15}{5}=3\)
b: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{\sqrt{x}+4}{x-2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{\sqrt{x}-2-\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{6}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)-6\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x\left(x-2\sqrt{x}+3\sqrt{x}-6\right)-6\sqrt{x}-6}{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
\(=\dfrac{x^2+x\sqrt{x}-6x-6\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\cdot\sqrt{x}}\)
