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Điều kiện xác định : \(0< x\ne1\)
\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
\(P=\left(\frac{1}{\sqrt{a}}+\frac{\sqrt{a}}{\sqrt{a}+1}\right):\left(\frac{\sqrt{a}}{a+\sqrt{a}}\right)\)
\(=\frac{\sqrt{a}+1+a}{\sqrt{a}\left(\sqrt{a}+1\right)}:\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}\right)=\frac{\left(a+\sqrt{a}+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}=\frac{a+\sqrt{a}+1}{\sqrt{a}}\)
\(a=\frac{2}{\sqrt{5}-1}-\frac{2}{\sqrt{5}+1}=\frac{2\sqrt{5}+2-2\sqrt{5}+2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{4}{4}=1\)
\(\Rightarrow P=3\)
\(P=\frac{a+\sqrt{a}+1}{\sqrt{a}}=\sqrt{a}+\frac{1}{\sqrt{a}}+1\ge2\sqrt{\frac{\sqrt{a}}{\sqrt{a}}}+1=3\)
\(\Rightarrow P_{min}=3\) khi \(a=1\)
\(ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)
\(P=\frac{2a+4}{a\sqrt{a}-1}+\frac{\sqrt{a}+2}{a+\sqrt{a}+1}-\frac{2}{\sqrt{a}-1}\)
\(=\frac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\frac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\frac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)
Ta có:
\(P=\frac{2a+4}{a\sqrt{a}-1}+\frac{\sqrt{a}+2}{a+\sqrt{a}+1}-\frac{2}{\sqrt{a}-1}\)
\(P=\frac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(P=\frac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(P=\frac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)
\(P=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)