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a,Với \(a>0;a\ne1\)
\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{\sqrt{a}-1+a-\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)^2}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{a-1}{a+\sqrt{a}}\)
b, Ta có : \(1=\frac{a+\sqrt{a}}{a+\sqrt{a}}\)mà \(a-1=\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\)
\(a+\sqrt{a}=\sqrt{a}\left(\sqrt{a}+1\right)\)vì \(\sqrt{a}-1< \sqrt{a}\)
Vậy \(\frac{a-1}{a+\sqrt{a}}< 1\)hay \(M< 1\)
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a-1}}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-2\sqrt{a}+1}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
Mà \(\sqrt{a}-1< \sqrt{a}\) => \(\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Vậy M < 1.
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}=\frac{\sqrt{a}}{\sqrt{a}}-\frac{1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)
Với \(x>0;x\ne1\)
\(M=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)
\(=1-\frac{1}{\sqrt{a}}< 1\)hay M < 1
\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}\right)^2-2\sqrt{a}+1}\)
\(=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(=\frac{\sqrt{a}-1}{\sqrt{a}}\)
Mà : \(\sqrt{a}-1< \sqrt{a}\Rightarrow\frac{\sqrt{a}-1}{\sqrt{a}}< 1\)
Vậy \(M< 1\)