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So sánh:
\(P=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\)
\(Q=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
Ta có : \(P=\left\{\frac{4}{7}+5+\frac{5}{7^3}\right\}+\left\{\frac{3}{7^2}+\frac{6}{7^4}\right\}\)
\(Q=\left\{\frac{4}{7}+5+\frac{5}{7^3}\right\}+\left\{\frac{5}{7^4}+\frac{6}{7^2}\right\}\)
So sánh : \(\frac{3}{7^2}+\frac{6}{7^4}\)và \(\frac{5}{7^4}+\frac{6}{7^2}\)
Ta có : \(\frac{3}{7^2}+\frac{6}{7^4}=\frac{49.3}{7^4}+\frac{6}{7^4}\)
\(\frac{5}{7^4}+\frac{6}{7^2}=\frac{5}{7^4}+\frac{49.6}{7^4}\)
Vì 49.3 + 6 < 49.6 + 5 nên Q > P.
:V Làm sai hết rồi sai ngay từ bước đầu tiên.
\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)
\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{3}{20}\)
\(=\frac{-11}{12}\)
\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)
= \(-\frac{7}{30}\)
Ta có :
\(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\)\(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(=\)\(\frac{2}{7}-\frac{1}{\frac{7}{2}}\)
\(=\)\(\frac{2}{7}-\frac{2}{7}\)
\(=\)\(0\)
Chúc bạn học tốt ~
\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)
\(=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\)