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\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+....+\dfrac{3}{59.61}\)
\(S=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+......+\dfrac{1}{59}-\dfrac{1}{61}\)
\(S=\left(\dfrac{1}{5}-\dfrac{1}{7}\right)+\left(\dfrac{1}{7}-\dfrac{1}{9}\right)+...+\left(\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{1}{5}-\dfrac{1}{61}\)
\(S=\dfrac{56}{305}\)
Vậy S = \(\dfrac{56}{305}\)
\(S=\dfrac{3}{5.7}+\dfrac{3}{7.9}+...+\dfrac{3}{59.61}\)
\(S=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(S=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}.\dfrac{56}{305}=\dfrac{84}{305}\)
a) \(\dfrac{4}{7}=\dfrac{4\cdot9}{7\cdot9}=\dfrac{36}{63}\)
\(\dfrac{13}{9}=\dfrac{13\cdot7}{9\cdot7}=\dfrac{91}{63}\)
\(\dfrac{8}{21}=\dfrac{8\cdot3}{21\cdot3}=\dfrac{24}{63}\)
b) \(\dfrac{1}{-36}=\dfrac{1\cdot5}{-36\cdot5}=\dfrac{-5}{180}\)
\(\dfrac{-8}{45}=\dfrac{-8\cdot4}{45\cdot4}=\dfrac{-32}{180}\)
\(\dfrac{13}{90}=\dfrac{13\cdot2}{90\cdot2}=\dfrac{26}{180}\)
c) \(3=\dfrac{3}{1}=\dfrac{3\cdot23}{1\cdot23}=\dfrac{69}{23}\)
\(-1=\dfrac{-1}{1}=\dfrac{-1\cdot23}{1\cdot23}=\dfrac{-23}{23}\)
\(\dfrac{17}{23}\) giữ nguyên
a, Ta có: \(\dfrac{32}{37}>\dfrac{32}{54}>\dfrac{19}{54}\Rightarrow\dfrac{32}{37}>\dfrac{19}{54}\)
b, Ta có: \(\dfrac{18}{53}>\dfrac{18}{54}=\dfrac{1}{3}\Rightarrow\dfrac{18}{53}>\dfrac{1}{3}\left(1\right)\)
\(\dfrac{26}{78}=\dfrac{1}{3}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{18}{53}>\dfrac{26}{78}\)
c, Ta thấy: \(\dfrac{25}{103}< \dfrac{25}{100}=\dfrac{1}{4}\left(1\right)\)
\(\dfrac{74}{295}>\dfrac{74}{296}=\dfrac{1}{4}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{25}{103}< \dfrac{74}{295}\)
A =\(\dfrac{4}{2.5}+\dfrac{4}{5.8}+\dfrac{4}{8.11}+...+\dfrac{4}{65.68}\)
A = \(\dfrac{4}{3}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{65.68}\right)\)
A = \(\dfrac{4}{3}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{65}-\dfrac{1}{68}\right)\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-...-\left(\dfrac{1}{65}-\dfrac{1}{65}\right)-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-0-0-0-...-0-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\left[\dfrac{1}{2}-\dfrac{1}{68}\right]\)
A = \(\dfrac{4}{3}.\dfrac{33}{68}\)
A = \(\dfrac{11}{17}\)
a) Để phân số \(\dfrac{12}{n}\) có giá trị nguyên thì :
\(12⋮n\)
\(\Leftrightarrow n\inƯ\left(12\right)\)
\(\Leftrightarrow n\in\left\{-1;1;-12;12;-2;2;-6;6;-3;3;-4;4\right\}\)
Vậy \(n\in\left\{-1;1;-12;12;-2;2-6;6;-3;3;-4;4\right\}\) là giá trị cần tìm
b) Để phân số \(\dfrac{15}{n-2}\) có giá trị nguyên thì :
\(15⋮n-2\)
\(\Leftrightarrow x-2\inƯ\left(15\right)\)
Tới đây tự lập bảng zồi làm típ!
c) Để phân số \(\dfrac{8}{n+1}\) có giá trị nguyên thì :
\(8⋮n+1\)
\(\Leftrightarrow n+1\inƯ\left(8\right)\)
Lập bảng rồi làm nhs!
Gọi phân số tối giản cần tìm là \(\dfrac{a}{b}\)
Ta có:\(\dfrac{a}{b}\):\(\dfrac{5}{11}\)=\(\dfrac{11a}{5b}\)
\(\dfrac{a}{b}\):\(\dfrac{11}{21}\)\(\dfrac{21a}{11b}\)
\(\dfrac{a}{b}\):\(\dfrac{25}{28}\)=\(\dfrac{28a}{25b}\)
Vì cả 3 thương trên là số tự nhiên nên a chia hết cho 5,11,25\(\)\(\Rightarrow\)a\(\in\)BCNN(5;11;25)\(\Rightarrow\)a=275
Do đó b\(\in\)ƯCLN(11,21,28)=1
Vậy phân số tối giản cần tìm là \(\dfrac{275}{1}\)
S = \(\dfrac{3}{1.2}\)+\(\dfrac{3}{2.3}\)+\(\dfrac{3}{3.4}\)+\(\dfrac{3}{4.5}\)+...+\(\dfrac{3}{2015.2016}\)
= 3.\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2015.2016}\right)\)
= 3.\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\right)\)
= 3.\(\left(1-\dfrac{1}{2016}\right)\) = 3.\(\dfrac{2015}{2016}\)=\(\dfrac{3.2015}{2016}\)=\(\dfrac{1.2015}{672}\)=\(\dfrac{2015}{672}\)
Vậy S = \(\dfrac{2015}{672}\)
Ta có S=\(\dfrac{3}{1.2}+\dfrac{3}{2.3}+\dfrac{3}{3.4}+\dfrac{3}{4.5}+...+\dfrac{3}{2015.2016}\)
=3.(\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2015.2016}\))
=3.(\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2015}-\dfrac{1}{2016}\))
=\(3.\left(1-\dfrac{1}{2016}\right)\)
= \(3-\dfrac{1}{672}\)=\(\dfrac{2015}{672}=2\dfrac{671}{672}\)
A. \(\dfrac{7}{11}\)