a , Khi x = 25 thì \(P=\dfrac{2\sqrt{25}+1}{\sqrt{25}}=\dfrac{10+1}{5}=\dfrac{11}{5}\)
b, Ta có : \(Q=\dfrac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}-2}=\dfrac{x-3\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
c, Ta có : \(M=\dfrac{Q}{P}=\dfrac{\dfrac{\sqrt{x}-2}{\sqrt{x}}}{\dfrac{2\sqrt{x}+1}{\sqrt{x}}}=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\)
Do \(|M|\ge M\) nên x thỏa mãn mọi số nguyên tố khác 2
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