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\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n-1}\right)}\)\(A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{n}-\sqrt{n-1}\)
\(A=\sqrt{n}-\sqrt{1}\)
\(B=\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}-\sqrt{2}\right)\left(\sqrt{1}+\sqrt{2}\right)}+\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{\sqrt{24}+\sqrt{25}}{\left(\sqrt{24}-\sqrt{25}\right)\left(\sqrt{24}+\sqrt{25}\right)}\)
\(B=-\left(\sqrt{1}+\sqrt{2}\right)-\left(\sqrt{2}+\sqrt{3}\right)-...-\sqrt{24}+\sqrt{25}\)
\(B=-1-2\sqrt{2}-2\sqrt{3}-...-\sqrt{24}-5\)
\(B=-1-2\sqrt{2}-2\sqrt{3}-...-\sqrt{24}-5\)
\(B=-6-2\sqrt{2}-2\sqrt{3}-...-2\sqrt{24}\)
ta có \(\frac{1}{\sqrt{1}+\sqrt{2}}=\frac{\sqrt{1}-\sqrt{2}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{1}-\sqrt{2}\right)}=\frac{\sqrt{1}-\sqrt{2}}{1-2}=\sqrt{1}-\sqrt{2}\)
mấy cái kia cũng thế a
\(=>A=\left(\sqrt{2}-1\right)+\left(\sqrt{3}-2\right)+...+\left(\sqrt{n}-\sqrt{n-1}\right)\)=>A= căn n -1
\(A=\frac{\sqrt{1}+\sqrt{2}}{1-2}-\frac{\sqrt{2}+\sqrt{3}}{2-3}+\frac{\sqrt{3}+\sqrt{4}}{3-4}-...-\frac{\sqrt{24}+\sqrt{25}}{24-25}\)
\(=-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{24}+\sqrt{25}\)
\(=-\sqrt{1}+\sqrt{25}\)
\(=-1+5\)
\(=4.\)
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\frac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(\Rightarrow\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{25\sqrt{24}+25\sqrt{24}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{24}}-\frac{1}{\sqrt{25}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{25}}=1-\frac{1}{5}=\frac{4}{5}\)
với n >0, ta có :
\(\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)=n+1-n=1\Rightarrow\frac{1}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}+\sqrt{n}\)
Gọi biểu thức đã cho là A
\(A=\frac{1}{-\left(\sqrt{2}-\sqrt{1}\right)}-\frac{1}{-\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{1}{-\left(\sqrt{8}-\sqrt{7}\right)}-\frac{1}{-\left(\sqrt{9}-\sqrt{8}\right)}\)
\(A=-\frac{1}{\sqrt{2}-\sqrt{1}}+\frac{1}{\sqrt{3}-\sqrt{2}}-...-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{9}-\sqrt{8}}\)
\(A=-\left(\sqrt{2}+\sqrt{1}\right)+\left(\sqrt{3}+\sqrt{2}\right)-...-\left(\sqrt{8}+\sqrt{7}\right)+\left(\sqrt{9}+\sqrt{8}\right)\)
\(A=-\sqrt{1}+\sqrt{9}=2\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\frac{2\sqrt{3}+\sqrt{18}+2\sqrt{3}-\sqrt{18}}{4-6}\right)-\frac{1}{\sqrt{2}}.\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}.\left(2\sqrt{3}\right)-\frac{1}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}-\frac{2\sqrt{6}-6}{\sqrt{2}+1}-\frac{1}{\sqrt{2}}\)
Nhân liên hiệp ta được :
\(\frac{\sqrt{1}+\sqrt{2}}{\left(\sqrt{1}-\sqrt{2}\right)\left(\sqrt{1}+\sqrt{2}\right)}+\frac{\sqrt{2}+\sqrt{3}}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+...+\frac{\sqrt{24}+\sqrt{25}}{\left(\sqrt{24}-\sqrt{25}\right)\left(\sqrt{24}+\sqrt{25}\right)}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{1-2}+\frac{\sqrt{2}+\sqrt{3}}{2-3}+...+\frac{\sqrt{24}+\sqrt{25}}{24-25}\)
\(=-\sqrt{1}-\sqrt{2}-\sqrt{2}-\sqrt{3}-....-\sqrt{24}-\sqrt{25}\)
\(=-\left[\frac{\left(\sqrt{25}+\sqrt{1}\right).25}{2}+\frac{\left(\sqrt{24}+\sqrt{2}\right).23}{2}\right]\)
\(=...\)