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cho mình nghi lại nha mn \(\dfrac{-3}{5}+\dfrac{28}{5}\cdot\left(\dfrac{43}{56}+\dfrac{5}{24}-\dfrac{21}{63}\right)\)
\(=\dfrac{7}{12}-\dfrac{3}{14}=\dfrac{98-36}{168}=\dfrac{62}{168}=\dfrac{31}{84}\)
\(C=70.\left(131313.\left(\dfrac{1}{565656}+\dfrac{1}{727272}+\dfrac{1}{909090}\right)\right)\)
\(C=70.\left(131313.\dfrac{1}{235690}\right)\)
\(C=70.\dfrac{39}{70}\)
\(C=39\)
\(27\cdot\left(\dfrac{373737}{424242}+\dfrac{373737}{565656}+\dfrac{373737}{727272}\right)\\ =27\cdot\left(\dfrac{37}{42}+\dfrac{37}{56}+\dfrac{37}{72}\right)\\ =27\cdot37\cdot\left(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\\ =999\cdot\left(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\\ =999\cdot\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\\ =999\cdot\left(\dfrac{1}{6}-\dfrac{1}{9}\right)\\ =999\cdot\dfrac{1}{18}\\ =\dfrac{111}{2}\)
Bài 1:
1: \(17A=\dfrac{17^{19}+17}{17^{19}+1}=1+\dfrac{16}{17^{19}+1}\)
\(17B=\dfrac{17^{18}+17}{17^{18}+1}=1+\dfrac{16}{17^{18}+1}\)
mà \(17^{19}+1>17^{18}+1\)
nên 17A>17B
hay A>B
2: \(C=\dfrac{98^{99}+98^{10}+1-98^{10}}{98^{89}+1}=98^{10}+\dfrac{1-98^{10}}{98^{89}+1}\)
\(D=\dfrac{98^{98}+98^{10}+1-98^{10}}{98^{88}+1}=98^{10}+\dfrac{1-98^{10}}{98^{88}+1}\)
mà \(98^{89}+1>98^{88}+1\)
nên C>D
a)Ta có : \(\dfrac{1212}{1414}=\dfrac{12.101}{14.101}=\dfrac{12}{14}\)
\(\dfrac{121212}{242424}=\dfrac{12.10101}{24.10101}=\dfrac{12}{24}\)
Vậy \(\dfrac{12}{24}=\dfrac{1212}{2424}=\dfrac{121212}{242424}\)
Ta có : \(\dfrac{2424}{3535}=\dfrac{24.101}{35.101}=\dfrac{24}{35}\)
\(\dfrac{242424}{353535}=\dfrac{24.10101}{35.10101}=\dfrac{24}{35}\)
Vậy \(\dfrac{24}{35}=\dfrac{2424}{3535}=\dfrac{242424}{353535}\)
\(18:\left(\dfrac{1111}{3636}+\dfrac{141414}{727272}-\dfrac{808}{1818}\right)\)
=18:\(\left(\dfrac{11}{36}+\dfrac{7}{36}-\dfrac{4}{9}\right)=18:\left(\dfrac{11+7-16}{36}\right)\)
=18:\(\dfrac{1}{18}\)
=324
TICK hộ nhé các bạn