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\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
Lời giải:
a)
\(a=\sqrt{2+\sqrt{3}}=\sqrt{\frac{4+2\sqrt{3}}{2}}=\sqrt{\frac{(\sqrt{3}+1)^2}{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}=b\)
b)
\( b=\sqrt{5-\sqrt{12+1+2\sqrt{12}}}=\sqrt{5-\sqrt{(\sqrt{12}+1)^2}}\)
\(=\sqrt{5-(\sqrt{12}+1)}=\sqrt{4-\sqrt{12}}\)
\(=\sqrt{4-2\sqrt{3}}=\sqrt{3+1-2\sqrt{3}}=\sqrt{(\sqrt{3}-1)^2}=\sqrt{3}-1=c\)
c)
\(\sqrt{n+2}>\sqrt{n+1}; \sqrt{n+1}> -\sqrt{n}\)
\(\Rightarrow \sqrt{n+2}+\sqrt{n+1}> \sqrt{n+1}-\sqrt{n}\)
A.\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\) \(=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)\left(n+1-n\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}\)
=\(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
b. ap dungtinh B =\(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
