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a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)
=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)
=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)
=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)
=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)
=\(\dfrac{36}{208}-\dfrac{5}{3}\)
=\(\dfrac{108}{624}-\dfrac{1040}{624}\)
=\(\dfrac{-932}{624}\)
=\(\dfrac{233}{156}\)
còn câu b mk chưa học nên mk chịu
Giải:
5/9:13/7+5/9:13/9 -1 2/3
=5/9.7/13+5/9.9/13-5/3
=5/9.(7/13+9/13)-5/3
=5/9.16/13-5/3
=80/117-5/3
=-115/117
4 2/5 : 0,5% -1 3/7 .14% +(-0,5)
=22/5:1/200-10/7.7/50 +(-1/2)
=880-1/5-1/2
=8793/10
`[-5]/13 . 2/11+[-5]/11 . 9/13+1`
`=[-5]/13 . 2/11+[-5]/13 . 9/11+1`
`=[-5]/13 . (2/11+9/11)+1`
`=[-5]/13 . 11/11+1`
`=[-5]/13+1=[-5]/13+13/13=8/13`
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
\(\dfrac{1}{4}\cdot\dfrac{1}{3}+\dfrac{1}{6}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{7}\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{7}\right)\\ =\dfrac{1}{4}\cdot\dfrac{9}{14}\\ =\dfrac{9}{56}\)
Ngô Hải Nam ơi bn trả lời giúp mik ik
bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên
a) \(4,5:\left[\left(\dfrac{9-10}{6}\right)-\dfrac{9}{5}+\dfrac{12}{5}\right]-\dfrac{1}{7}\)
\(=4,5:\left(\dfrac{-1}{6}-\dfrac{-3}{5}\right)-\dfrac{1}{7}\)
=\(4,5:\left(\dfrac{-5+18}{30}\right)-\dfrac{1}{7}\)
=\(4,5:\dfrac{13}{30}-\dfrac{1}{7}\)=\(\dfrac{135}{13}-\dfrac{1}{7}=\dfrac{932}{91}\)
b) \(\dfrac{13}{3}:\left(\dfrac{1}{4}+\dfrac{5}{4}\right)-\dfrac{20}{3}\)
=\(\dfrac{13}{3}.\dfrac{2}{3}-\dfrac{20}{3}\)=\(\dfrac{26}{9}-\dfrac{20}{3}=\dfrac{26}{9}-\dfrac{60}{9}=\dfrac{-34}{9}\)
c) \(5.\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+.....+\dfrac{1}{91.94}\right)\)
\(=5.\left[\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{91}-\dfrac{1}{94}\right)\right]\)
\(=5.\left[\dfrac{1}{3}.\left(1-\dfrac{1}{94}\right)\right]\)
=\(5.\left(\dfrac{1}{3}.\dfrac{93}{94}\right)\)
\(=5.\dfrac{31}{94}=\dfrac{155}{94}\)
Chúc bạn học tốt 
Sao số nào cũng lớn hơn \(\dfrac{1}{5}\) thế bạn nhỉ??