a: Thay x=25 vào A, ta được:
\(A=\dfrac{25+3}{\sqrt{25}-2}=\dfrac{28}{5-2}=\dfrac{28}{3}\)
b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+7}{x-\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}+1\right)+\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-2\sqrt{x}-3\sqrt{x}-3+\sqrt{x}+7}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
c: \(P=A\cdot B=\dfrac{x+3}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{x+3}{\sqrt{x}+1}\)
\(P=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)
\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=4\)
=>\(P=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2>=4-2=2\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi \(\sqrt{x}+1=\sqrt{4}=2\)
=>x=1(nhận)
