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\(A=\dfrac{4116-14}{10290-35}=\dfrac{14\times\left(294-1\right)}{35\times\left(294-1\right)}=\dfrac{14}{35}\)
\(B=\dfrac{2929-101}{2\times1919+404}=\dfrac{101\times\left(29-1\right)}{101\times\left(38+4\right)}=\dfrac{29-1}{38+4}=\dfrac{28}{32}=\dfrac{7}{8}\)
b: \(=\dfrac{50\left(62+44\cdot2\right)}{38\left(27+73\right)}=\dfrac{50\cdot150}{38\cdot100}=\dfrac{75}{38}\)
c: \(=\dfrac{7112}{10255}=\dfrac{1016}{1465}\)
a)\(\dfrac{17}{15}>1;\dfrac{29}{37}< 1\Leftrightarrow\dfrac{17}{15}>\dfrac{29}{37}\)
b) \(\dfrac{13}{17}>\dfrac{13}{18}\Leftrightarrow\dfrac{13}{17}>\dfrac{12}{18}\)
d)\(1-\dfrac{2017}{2018}=\dfrac{1}{2018}\)
\(1-\dfrac{2018}{2019}=\dfrac{1}{2019}\)
\(\dfrac{1}{2018}>\dfrac{1}{2019}\Leftrightarrow\dfrac{2017}{2018}< \dfrac{2018}{2019}\)
e) \(\dfrac{2018}{2017}< 1;\dfrac{2019}{2018}>1\Leftrightarrow\dfrac{2018}{2017}< \dfrac{2019}{2018}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(-\dfrac{1}{11}+\dfrac{1}{11}\right)+\left(-\dfrac{1}{12}+\dfrac{1}{12}\right)+\left(-\dfrac{1}{13}+\dfrac{1}{13}\right)+\left(-\dfrac{1}{14}+\dfrac{1}{14}\right)+\left(-\dfrac{1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\\ =\dfrac{1}{16}\)
Tính nhanh :
\(\dfrac{1}{10}+\dfrac{-1}{11}+\dfrac{1}{12}+\dfrac{-1}{13}+\dfrac{1}{14}+\dfrac{-1}{15}+\dfrac{1}{16}+\dfrac{-1}{10}+\dfrac{1}{11}+\dfrac{-1}{12}+\dfrac{1}{13}+\dfrac{-1}{14}+\dfrac{1}{15}\)
\(=\left(\dfrac{1}{10}+\dfrac{-1}{10}\right)+\left(\dfrac{-1}{11}+\dfrac{1}{11}\right)+\left(\dfrac{1}{12}+\dfrac{-1}{12}\right)+\left(\dfrac{-1}{13}+\dfrac{1}{13}\right)+\left(\dfrac{1}{14}+\dfrac{-1}{14}\right)\)
\(+\left(\dfrac{-1}{15}+\dfrac{1}{15}\right)+\dfrac{1}{16}\)
\(=0+0+...+0+\dfrac{1}{16}\)
\(=\dfrac{1}{16}\)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
a: 51/56=1-5/56
61/66=1-5/66
mà -5/56<-5/66
nên 51/56<61/66
b: 41/43<1<172/165
c: \(\dfrac{101}{506}>0>-\dfrac{707}{3534}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK
a) \(\dfrac{4116-14}{10290-35}=\dfrac{4102}{10255}=\dfrac{2051.2}{2051.5}=\dfrac{2}{5}\)
b) \(\dfrac{2929-101}{2.1919+404}=\dfrac{2929-101}{3838+404}=\dfrac{2828}{4242}=\dfrac{1414.2}{1414.3}=\dfrac{2}{3}\)
Rút gọn :
A= \(\dfrac{4116-14}{10290-35}\)= \(\dfrac{4102}{10255}=\dfrac{4102:2051}{10255:2051}=\dfrac{2}{5}\)
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
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