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\(M=\dfrac{40^{20}-2^{20}+6^{20}}{6^{20}-3^{20}+9^{20}}\)
\(=\dfrac{\left(2^3.5\right)^{20}-2^{20}+\left(2.3\right)^{20}}{\left(2.3\right)^{20}-3^{20}+\left(3^2\right)^{20}}\)
\(=\dfrac{2^{60}.5^{20}-2^{20}+2^{20}.3^{20}}{2^{20}.3^{20}-3^{20}+3^{20}}\)
\(=\dfrac{2^{20}\left(2^{30}.5^{20}-1+3^{20}\right)}{3^{20}\left(2^{20}-1+3^{20}\right)}\)
có nhầm đề k nhỉ ?
\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
CHÚC BẠN HỌC TỐT NHA!!!
\(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{20}\)
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{19}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{20}}\right)\)
\(A=1-\dfrac{1}{2^{20}}=\dfrac{2^{20}-1}{2^{20}}\)
Chọn A
\(1,25:\dfrac{15}{20}+\left(25\%-\dfrac{5}{6}\right):4\dfrac{2}{3}\)
\(=\dfrac{5}{4}.\dfrac{4}{3}+\left(\dfrac{1}{4}-\dfrac{5}{6}\right).\dfrac{3}{14}\)
\(=\dfrac{5}{3}+\dfrac{7}{12}.\dfrac{3}{14}\)
\(=\dfrac{5}{3}+-\dfrac{1}{8}\)
\(=\dfrac{37}{24}\)
\(1,25:\dfrac{15}{20}+\left(25\%-\dfrac{5}{6}\right):4\dfrac{2}{3}\)
\(=\dfrac{5}{4}:\dfrac{3}{4}+\left(\dfrac{1}{4}-\dfrac{5}{6}\right):\dfrac{14}{3}\)
\(=\dfrac{5}{4}.\dfrac{4}{3}+\left(\dfrac{-7}{12}\right).\dfrac{3}{14}\)
= \(\dfrac{5}{3}-\dfrac{1}{8}=\dfrac{37}{24}\)
e: \(=\dfrac{5^{30}\cdot3^{20}}{3^{15}\cdot5^{30}}=3^5=243\)
Ta có :
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(M=\dfrac{20}{8\times14}+\dfrac{20}{14\times20}+\dfrac{20}{20\times26}+\dfrac{20}{26\times32}\)
\(\Rightarrow\dfrac{3}{10}M=\dfrac{6}{8\times14}+\dfrac{6}{14\times20}+\dfrac{6}{20\times26}+\dfrac{6}{26\times32}\)
\(\dfrac{3}{10}M=\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\)
\(\dfrac{3}{10}M=\dfrac{1}{8}-\dfrac{1}{32}=\dfrac{3}{32}\)
\(\Rightarrow M=\dfrac{3}{32}\div\dfrac{3}{10}=\dfrac{5}{16}\)
\(M=\dfrac{20}{112}+\dfrac{20}{280}+\dfrac{20}{520}+\dfrac{20}{832}\)
\(M=20.\left(\dfrac{1}{112}+\dfrac{1}{280}+\dfrac{1}{520}+\dfrac{1}{832}\right)\)
\(M=20.\left(\dfrac{1}{8.14} +\dfrac{1}{14.20}+\dfrac{1}{20.26}+\dfrac{1}{26.32}\right)\)
\(\Rightarrow6M=20.\left(\dfrac{6}{8.14}+\dfrac{6}{14.20}+\dfrac{6}{20.26}+\dfrac{6}{26.32}\right)\)
\(\Rightarrow6M=20.\left(\dfrac{1}{8}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{32}\right)\)
\(\Rightarrow6M=20.\left(\dfrac{1}{8}-\dfrac{1}{32}\right)\)
\(\Rightarrow6M=20.\dfrac{3}{32}\)
\(\Rightarrow6M=\dfrac{15}{8}\)
\(\Rightarrow M=\dfrac{15}{8}:6\)
\(\Rightarrow M=\dfrac{5}{16}\)
Vậy \(M=\dfrac{5}{16}\)