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Đặt A
Ta có công thức :
\(\frac{a}{b.c}=\frac{a}{c-b}.\left(\frac{1}{b}-\frac{1}{c}\right)\)
Dựa vào công thức, ta có
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)=\frac{5}{2}.\left(\frac{12}{25}\right)=\frac{6}{5}\)
Ai thấy đúng thì ủng hộ nha !!!
a, \(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
=\(\frac{5}{2}\left(\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{48.50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)=\(\frac{5}{2}.\frac{12}{25}\)=\(\frac{6}{5}\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{1}{2}.\frac{49}{100}=\frac{49}{200}\)
\(\frac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^9\cdot6^{14}-7\cdot2^{29}\cdot27^6}\)
\(=\frac{5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9}{5\cdot2^9\cdot\left(2\cdot3\right)^{14}-7\cdot2^{29}\cdot\left(3^3\right)^6}\)
\(=\frac{5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}}{5\cdot2^9\cdot2^{14}\cdot3^{14}-7\cdot2^{29}\cdot3^{18}}\)
\(=\frac{5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}}{5\cdot2^{23}\cdot3^{14}-7\cdot2^{29}\cdot3^{18}}\)
\(=\frac{2^{29}\cdot3^{18}\cdot\left(5\cdot2\cdot1-1\cdot3^2\right)}{2^{23}\cdot3^{14}\cdot\left(5\cdot1\cdot1-7\cdot2^6\cdot3^4\right)}\)
\(=\frac{2^6\cdot3^4\cdot1}{5-36288}\)
\(=\frac{5184}{-36283}=-\frac{5184}{36283}\)
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.2^{30}.3^{18}-2^2.3^{20}.2^{27}}{5.2^{10}.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{18}-2^{29}.3^{20}}{5.2^{29}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{29}.3^{18}\left(5.3-7\right)}\)
\(=\frac{10-9}{15-7}\)
\(=\frac{1}{8}\)
Đặt \(A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)
Ủng hộ mk nha !!! *_*
\(\text{Đ}\text{ặt}:A=\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+..+\frac{5}{100.102}\)
\(\frac{2}{5}A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{3}{6.8}+...+\frac{2}{100.102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{100}-\frac{1}{102}\)
\(\frac{2}{5}A=\frac{1}{2}-\frac{1}{102}\)
\(A=\frac{25}{51}:\frac{2}{5}\)
\(A=\frac{125}{102}\)