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Sử dụng liên hợp để trục căn thức ở mẫu:
\(\frac{1}{\sqrt{1}+\sqrt{5}}=\frac{\sqrt{5}-1}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}=\frac{\sqrt{5}-1}{5-1}=\frac{\sqrt{5}-1}{4}\)
Tương tự như vậy ta sẽ có:
\(N=\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\sqrt{13}-\sqrt{9}}{\left(\sqrt{13}-\sqrt{9}\right)\left(\sqrt{13}+\sqrt{9}\right)}+\frac{\sqrt{17}-\sqrt{13}}{\left(\sqrt{17}-\sqrt{13}\right)\left(\sqrt{17}+\sqrt{13}\right)}\)
\(+\frac{\sqrt{21}-\sqrt{17}}{\left(\sqrt{21}-\sqrt{17}\right)\left(\sqrt{21}+\sqrt{17}\right)}+\frac{\sqrt{25}-\sqrt{23}}{\left(\sqrt{25}-\sqrt{23}\right)\left(\sqrt{25}+\sqrt{23}\right)}\)
\(=\frac{\sqrt{5}-1}{4}+\frac{\sqrt{13}-\sqrt{9}}{4}+\frac{\sqrt{17}-\sqrt{13}}{4}+\frac{\sqrt{21}-\sqrt{17}}{4}+\frac{\sqrt{25}-\sqrt{23}}{4}\)
\(=\frac{\sqrt{5}-1+\sqrt{13}-\sqrt{9}+\sqrt{17}-\sqrt{13}+\sqrt{21}-\sqrt{17}+\sqrt{25}-\sqrt{23}}{4}\)
\(=\frac{\sqrt{5}-1-\sqrt{9}+\sqrt{21}+\sqrt{25}-\sqrt{23}}{4}=\frac{\sqrt{5}-1-3+\sqrt{21}+5-\sqrt{23}}{4}=\frac{1+\sqrt{5}+\sqrt{21}-\sqrt{23}}{4}\)
Với n > 0 Ta có:
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}=\frac{\sqrt{n+1}+\sqrt{n}}{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)
\(=\sqrt{n+1}+\sqrt{n}\)
\(\Rightarrow\frac{1}{\sqrt{16}-\sqrt{15}}-\frac{1}{\sqrt{15}-\sqrt{14}}+...+\frac{1}{\sqrt{10}-\sqrt{9}}\)
\(=\sqrt{16}+\sqrt{15}-\sqrt{15}-\sqrt{14}+...+\sqrt{10}+\sqrt{9}\)
\(\sqrt{16}+\sqrt{9}=3+4=7\)
\(\frac{-63}{108}\)= \(\frac{-7}{12}\)
\(\frac{-33}{-77}\)= \(\frac{3}{7}\)
\(\frac{-5}{10}\)=\(\frac{-1}{2}\)
\(\frac{14}{63}\)=\(\frac{2}{9}\)
\(\frac{-15}{25}\)=\(\frac{-3}{5}\)
\(\frac{-45}{18}\)=\(\frac{-5}{2}\)
\(\frac{12}{15}\)=\(\frac{4}{5}\)
\(\frac{20}{25}\)=\(\frac{4}{5}\)
\(\frac{31}{12}\):Là phân số tối giản
t.i.c.k nha
Đặt A=\(\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
\(\frac{A}{2}=\frac{13}{42}-\frac{15}{56}+\frac{17}{72}-...+\frac{197}{9702}-\frac{199}{4950}\)
\(=\frac{6+7}{6.7}-\frac{7+8}{7.8}+\frac{8+9}{8.9}-...+\frac{98+99}{98.99}-\frac{99+100}{99.100}\)
\(=\frac{1}{7}+\frac{1}{6}-\frac{1}{8}-\frac{1}{7}+\frac{1}{9}+\frac{1}{8}-...+\frac{1}{99}+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}\)
\(=\frac{1}{6}-\frac{1}{100}=\frac{47}{300}\)
\(\Rightarrow A=\frac{47}{300}.2=\frac{47}{150}\)
\(\Rightarrow Q=\frac{85}{25}+\frac{9}{10}-\frac{11}{5}+\frac{47}{150}=\frac{181}{75}\)
Vậy Q=\(\frac{181}{75}\).