a: Thay x=2 vào A, ta được:
\(A=\dfrac{2+5}{\sqrt{2}+1}=\dfrac{7}{\sqrt{2}+1}=7\left(\sqrt{2}-1\right)=7\sqrt{2}-7\)
b: \(B=\dfrac{x-\sqrt{x}+4}{x-\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+4-\left(\sqrt{x}-2\right)^2-\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+4-x+4\sqrt{x}-4-x-\sqrt{x}}{\left(\sqrt[]{x}-2\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=-\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
c:
ĐKXĐ của C là x>0; x<>4
\(C=A:B=\dfrac{x+5}{\sqrt{x}+1}:\dfrac{-\sqrt{x}}{\sqrt{x}+1}=\dfrac{x+5}{-\sqrt{x}}\)
\(=-\dfrac{x+5}{\sqrt{x}}=-\left(\sqrt{x}+\dfrac{5}{\sqrt{x}}\right)\)
\(\sqrt{x}+\dfrac{5}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{5}{\sqrt{x}}}=2\sqrt{5}\) với mọi x thỏa mãn ĐKXĐ
=>\(C=-\left(\sqrt{x}+\dfrac{5}{\sqrt{x}}\right)< =-2\sqrt{5}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=5
