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Câu 1 : \(1,321338308x10^{-4}\)
Câu 2 : \(1316,572106\)
Câu 3 : \(1,641302619x10^{-13}\)
Ủng hộ nhé,tớ đang âm.
\(\dfrac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}=\dfrac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9+3^{10}.2^{20}}\)
\(=\dfrac{2^{19}.3^9+3^9.5.2^{18}}{2^9.3^9+3^{10}.2^{20}}=\dfrac{3^9.2^{18}.\left(2+5\right)}{2^9.3^9.\left(1+6\right)}\)
\(=\dfrac{3^9.2^{18}}{2^9.3^9}=2^9=512\)
~ Học tốt ~
\(B=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+\dfrac{2}{4.5.6}+\dfrac{2}{5.6.7}+\dfrac{2}{6.7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{6.7}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{7.8}\)
\(=\dfrac{1}{2}-\dfrac{1}{56}=\dfrac{27}{56}\)
a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(A=5\left(1-\dfrac{1}{100}\right)\)
\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)
b, \(C=1.2.3+2.3.4+...+8.9.10\)
\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)
\(C=\dfrac{8.9.10.11}{4}=1980.\)
c, https://hoc24.vn/hoi-dap/question/384591.html
Câu này bạn vào đây mình đã giải câu tương tự nhé.
\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)
\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)
\(\Leftrightarrow A=\dfrac{99}{20}\)
\(B=5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
\(B=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right):\dfrac{7}{4}\)
\(B=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{63}{6}-\dfrac{14}{3}\right):\dfrac{7}{4}\)
\(B=\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{63}{6}-\dfrac{28}{6}\right):\dfrac{7}{4}\)
\(B=\dfrac{59}{10}:\dfrac{3}{2}-\dfrac{35}{6}:\dfrac{7}{4}\)
\(B=\dfrac{59}{10}.\dfrac{2}{3}-\dfrac{35}{6}.\dfrac{4}{7}\)
\(B=\dfrac{59}{15}-\dfrac{10}{3}\)
\(B=\dfrac{59}{15}-\dfrac{50}{15}\)
\(B=\dfrac{3}{5}\)
B=\(5\dfrac{9}{10}:\dfrac{3}{2}-\left(2\dfrac{1}{3}.4\dfrac{1}{2}-2.2\dfrac{1}{3}\right):\dfrac{7}{4}\)
B=\(\dfrac{59}{10}:\dfrac{3}{2}-\left(\dfrac{7}{3}.\dfrac{9}{2}-2.\dfrac{7}{3}\right):\dfrac{7}{4}\)
B=\(\dfrac{59}{15}-\left(\dfrac{21}{2}-\dfrac{14}{3}\right):\dfrac{7}{4}=\dfrac{59}{15}-\left(\dfrac{63-28}{6}\right):\dfrac{7}{4}\)
B=\(\dfrac{59}{15}-\dfrac{35}{6}:\dfrac{7}{4}\)
B=\(\dfrac{59}{15}-\dfrac{10}{3}=\dfrac{59-50}{15}\)
B=\(\dfrac{3}{5}\)
\(A=\dfrac{4}{1\cdot2}+\dfrac{4}{2\cdot3}+\dfrac{4}{3\cdot4}+...+\dfrac{4}{2022\cdot2023}\\ A=4\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2022\cdot2023}\right)\\ A=4\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\\ A=4\cdot\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)\\ A=4\cdot\dfrac{2022}{2023}\\ A=\dfrac{8088}{2023}\)
1. \(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}{4\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{9}-\dfrac{1}{11}\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
2. \(B=\dfrac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\dfrac{1}{5}\)
3.\(C=\dfrac{2^2.3^2.\text{4^2.5^2}.5^2}{1.2^2.3^2.4^2.5.6^2}=\dfrac{125}{36}\)
4.D=\(D=\left(\dfrac{4}{5}-\dfrac{1}{6}\right).\dfrac{4}{9}.\dfrac{1}{16}=\dfrac{19}{30}.\dfrac{1}{36}=\dfrac{19}{1080}\)
\(I=\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}=\frac{5.2^{30}.3^{27}-2^2.3^{20}.2^{27}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{5.2^{30}.3^{27}-3^{30}.2^{29}}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{27}.\left(5.2-3^3\right)}{2^{28}.3^{18}.\left(5.3-2.7\right)}\)
\(=\frac{2^{29}.3^{27}.-17}{2^{18}.3^{18}}\)
\(=\frac{2^9.3^9.-17}{1}\)
Ta có \(H=\frac{\left(3.4.2^{16}\right)}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3.4.2^{16}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3.2^{18}}{11.2^{35}-2^{36}}\)
\(=\frac{3.2^{18}}{2^{35}.\left(11-2\right)}\)
\(=\frac{3.2^{18}}{2^{35}.3^2}\)
\(=\frac{1}{2^{17}.3}\)
\(A=\frac{15.3^{11}+4.27^1}{9^7}\)
\(\Rightarrow A=\frac{3.5.3^{11}+4.3^{3^1}}{\left(3^2\right)^7}\)
\(\Rightarrow A=\frac{3^{12}.5+4.3^3}{3^{14}}\)
\(\Rightarrow A=\frac{3^3.\left(5.3^8+4.3^3\right)}{3^{14}}\)
\(\Rightarrow A=\frac{32805+4}{177147}\)
\(\Rightarrow A=\frac{32809}{177147}\)
chịu