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2 tháng 4 2018

=36+\(\sqrt{17}\)

=40.12310563

k nhé

1 tháng 10 2020

\(ĐKXĐ:\hept{\begin{cases}a>0\\a\ne1\end{cases}}\)

\(P=\frac{2a+4}{a\sqrt{a}-1}+\frac{\sqrt{a}+2}{a+\sqrt{a}+1}-\frac{2}{\sqrt{a}-1}\)

\(=\frac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\frac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\frac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)

1 tháng 10 2020

Ta có:

\(P=\frac{2a+4}{a\sqrt{a}-1}+\frac{\sqrt{a}+2}{a+\sqrt{a}+1}-\frac{2}{\sqrt{a}-1}\)

\(P=\frac{2a+4+\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-2\left(a+\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(P=\frac{2a+4+a+\sqrt{a}-2-2a-2\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(P=\frac{a-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\)

\(P=\frac{\sqrt{a}}{a+\sqrt{a}+1}\)

7 tháng 7 2018

\(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\Rightarrow\)\(\frac{A}{\sqrt{2}}=\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)

                   \(=\frac{2+\sqrt{3}}{2+\left(\sqrt{3}+1\right)}+\frac{2-\sqrt{3}}{2-\left(\sqrt{3}-1\right)}\)

                   \(=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)       

                   \(=\frac{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\left(2-\sqrt{3}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

                  \(=\frac{\sqrt{3}+1}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}+\frac{\sqrt{3}-1}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

                    \(=\frac{2\sqrt{3}}{2\sqrt{3}}=1\)

                          

7 tháng 9 2018

\(A=\frac{\sqrt{2}-1}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\)\(\Leftrightarrow A=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\) 

\(\Leftrightarrow A=\sqrt{n}-1\)

10 tháng 7 2017

a ĐK \(a>0\)và \(a\ne1\)

\(M=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right).\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

b. Ta có \(M-1=\frac{\sqrt{a}-1}{\sqrt{a}}-1=\frac{\sqrt{a}-1-\sqrt{a}}{\sqrt{a}}=\frac{-1}{\sqrt{a}}< 0\)

Vậy \(M< 1\)