a) \(x=25\left(tmđk\right)\)
\(\rightarrow\sqrt{x}=5\)
Thay vào A ta có :
\(A=\dfrac{25-5+1}{5-1}=\dfrac{21}{4}\)
b) \(B=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+2\left(\sqrt{x}-2\right)-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(B=\left(\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\right)\)
\(B=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(B=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
a) Thay x=25 vào A, ta được:
\(A=\dfrac{25-5+1}{5-1}=\dfrac{21}{4}\)
b) Ta có: \(B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\)
\(=\dfrac{x+4\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
