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Tối mk hk rồi
B = \(\dfrac{30.4^7.3^{29}-5.4^{15}.2^{12}}{54.6^{14}.9^7-12.8^5.7^5}\)
B=\(\dfrac{5.6.\left(2^2\right)^7.3^{29}-5.\left(2^2\right)^{15}.2^{12}}{9.5.\left(2.3\right)^{14}.\left(3^2\right)^7-\left(3.4\right).\left(2^3\right)^5.7^5}\)
B=\(\dfrac{5.\left(2.3\right).2^{14}.3^{19}-5.2^{30}.2^{12}}{3^2.5.2^{14}.3^{14}-3.4.2^{15}.7^5}\)
B=\(\dfrac{5.2^{15}.3^{20}-5.2^{30}.2^{12}}{5.2^{14}.3^{16}-3.2^{17}.7^5}\)
B=\(\dfrac{5.\left(2^{15}.3^{20}-2^{30}.2^{12}\right)}{2^{14}.\left(5.3^{16}-3.2^3.7^5\right)}\)
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+....+\dfrac{3}{59.61}\)
\(S=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{59}-\dfrac{1}{61}\)
\(S=\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{1}{5}-\dfrac{1}{61}\)
\(S=\dfrac{56}{305}\)
Vậy S = \(\dfrac{56}{305}\)
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
\(S=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}.\dfrac{56}{305}=\dfrac{84}{305}\)
Ta có: \(\dfrac{2}{3}=\dfrac{2\times8}{3\times8}=\dfrac{16}{24}\)
\(\dfrac{7}{8}=\dfrac{7\times3}{8\times3}=\dfrac{21}{24}\)
\(\dfrac{11}{24}=\dfrac{11}{24}\)
\(\dfrac{2}{3};\dfrac{7}{8};\dfrac{11}{24}\)
lần lượt bằng: \(\dfrac{16}{24};\dfrac{21}{24};\dfrac{11}{24}\)
S=\(\dfrac{1}{1}-\dfrac{1}{4} +...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\)
S=\(\dfrac{1}{1}-\dfrac{1}{100}\)
S=1-\(\dfrac{1}{100}\)
S=\(\dfrac{99}{100}\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{28}.3-9^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3-9^{15}}{2^2}\)
\(=\dfrac{33-9^{15}}{4}\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{2^2.3^{28}}\)
=\(\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
=\(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
=\(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\)
=\(\dfrac{3^{29}.8}{2^2.3^{28}}\)
=\(\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
=3.2
=6
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
Ta có :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......................+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
a) \(\dfrac{-23}{3}=7\dfrac{-2}{3}\)
b) \(123,45=\dfrac{12345}{100}\cdot100\%=123,45\%\)
a, \(-\dfrac{23}{3}=-7\dfrac{2}{3}.\) (cách đổi: bn chỉ cần lấy tử chia mẫu là sẽ ra (tử, mẫu là số tự nhiên; tử > mẫu)).
b, \(123,45=\dfrac{12345}{100}=123,45\%.\)
~ Học tốt nha bn!!! ~
\(S=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2017}\)
\(\dfrac{2}{7}S=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\)
\(\dfrac{2}{7}S=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2015}-\dfrac{1}{2017}\)
\(\dfrac{2}{7}S=\dfrac{1}{3}-\dfrac{1}{2017}\)
\(\dfrac{2}{7}S=\dfrac{2014}{6051}\)
\(S=\dfrac{4028}{42357}\)
\(S=\dfrac{7}{3.5}+\dfrac{7}{5.7}+\dfrac{7}{7.9}+...+\dfrac{7}{2015.2107}\)
\(S=\dfrac{7}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{2015.2017}\right)\)
\(S=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}...+\dfrac{1}{2015}-\dfrac{1}{2017}\right)\)
\(S=\dfrac{7}{2}\left(\dfrac{1}{3}-\dfrac{1}{2017}\right)\)
\(S=\dfrac{7}{2}.\dfrac{2014}{6051}\)
\(S=\dfrac{4028}{42357}\)