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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{91}-\frac{1}{93}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{93}\right)\)
\(=\frac{1}{2}.\frac{92}{93}\)
\(=\frac{46}{93}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{91.93}\)
\(=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{91}-\frac{1}{93}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{93}\right)\)
\(=\frac{1}{2}.\frac{92}{93}\)
\(=\frac{46}{93}\)
Ta có:
\(A=\frac{3}{1\cdot5}+\frac{3}{5\cdot10}+...+\frac{3}{100\cdot105}\)
\(=\frac{3}{5}\cdot\left(\frac{5}{1\cdot5}+\frac{5}{5\cdot10}+...+\frac{5}{100\cdot105}\right)\)
\(=\frac{3}{5}\cdot\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{105}\right)\)
\(=\frac{3}{5}\left(1-\frac{1}{105}\right)=\frac{3}{5}\cdot\frac{104}{105}=\frac{312}{525}\)
a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
:V Làm sai hết rồi sai ngay từ bước đầu tiên.
\(\frac{1}{3.4}-\frac{1}{4.5}-\frac{1}{5.6}-....-\frac{1}{9.10}\)
\(=\frac{1}{3.4}-\left(\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{9.10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\left(\frac{1}{4}-\frac{1}{10}\right)\)
\(=\frac{1}{12}-\frac{3}{20}\)
\(=\frac{-11}{12}\)
\(\frac{1}{3.4}-\frac{1}{4.5}-...-\frac{1}{9.10}\)
= \(-\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
= \(-\left(\frac{1}{3}-\frac{1}{10}\right)\)
= \(-\frac{7}{30}\)
đặt A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}\)
ta có:
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}\)
=> 2A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{91.93}\)
=> 2A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}\)
=> 2A = \(\frac{1}{1}-\frac{1}{93}\)
2A = \(\frac{92}{93}\)
=> A = \(\frac{92}{93}:2\)
A = \(\frac{46}{93}\)
=1-1/3+1/3-1/5+1/5-1/7+.........+1/91-1/93
=>1-1/93=92/92