\(\frac{11}{5^2.6^2}+\frac{13}{6^2.7^2}+\frac{15}{7^2.20^2}+...+\frac{39}{19^2.20^2}\)
Chứng minh dãy phép tính trên ko phải là số nguyên
K
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Những câu hỏi liên quan
10 tháng 11 2018
a. \(\frac{\left(-5\right)^2.20^4}{8^2.\left(-125\right)}=\frac{\left(-5\right)^2.5^4.2^8}{2^6.\left(-5\right)^3}=\left(-5\right)^3.2^2=\left(-125\right).4=-500\)
b, \(\frac{15^{11}.5^7.9^2}{5^{18}.27^6}=\frac{3^{11}.5^{11}.5^7.3^4}{5^{18}.3^{18}}=\frac{3^{15}.5^{18}}{5^{18}.3^{18}}=\frac{1}{3^3}=\frac{1}{27}\)
BT
28 tháng 4 2017
Ta có: \(\frac{3}{1^2.2^2}=\frac{3}{1.4}=1-\frac{1}{4}\); \(\frac{5}{2^2.3^2}=\frac{5}{4.9}=\frac{1}{4}-\frac{1}{9}\); \(\frac{7}{3^2.4^2}=\frac{7}{9.16}=\frac{1}{9}-\frac{1}{16}\); ...; \(\frac{39}{19^2.20^2}=\frac{39}{361.400}=\frac{1}{361}-\frac{1}{400}\)
Gọi tổng đó là A => A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{361}-\frac{1}{400}\)
=> \(A=1-\frac{1}{400}=\frac{399}{400}< \frac{400}{400}=1\)
=> A < 1
1 tháng 6 2018
b,\(D=2.\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{n.\left(n+2\right)}\right)\)
\(\Rightarrow D=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{n.\left(n+2\right)}\)
\(\Rightarrow D=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+2}\)
\(\Rightarrow D=1-\frac{1}{n+2}=\frac{n}{n+2}< \frac{n+2}{n+2}=1\left(1\right)\)
\(\Rightarrow D=\frac{n}{n+2}>0\left(2\right)\)
Từ (1);(2)\(\Rightarrow0< D< 1\)
\(\Rightarrowđpcm\)
F
20 tháng 7 2020
a,\(C>0\)
\(C=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{19}< 9;\frac{1}{11}< 1\)
\(\Rightarrow0< A< 1\)
\(\Rightarrow A\notinℤ\)
c,\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
Ta quy đồng 3 số đầu
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}>\frac{6.2}{12}=1\)
\(E=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}\)
\(=\frac{2}{6}+\frac{2}{8}+\frac{2}{10}+\frac{2}{7}+\frac{2}{9}+\frac{2}{11}< \frac{6.2}{6}=2\)
\(1< E< 2\)
\(E\notinℤ\)
LM
8 tháng 3 2017
a) Ta có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}>\frac{6}{19}.5=\frac{30}{19}>1\)
\(\Rightarrow S>1\)
Ta lại có:
\(\frac{6}{15}+\frac{6}{16}+...+\frac{6}{19}< \frac{6}{15}.5=\frac{30}{15}=2\)
\(\Rightarrow S< 2\)
Vậy, 1 < S < 2
b) \(1< S< 2\Rightarrow S\notin Z\)
NH
27 tháng 3 2019
\(a.\frac{108}{119}.\frac{107}{211}+\frac{108}{119}.\frac{104}{211}=\frac{108}{119}.\left(\frac{107}{211}+\frac{104}{211}\right)=\frac{108}{119}.1=108\)
S
23 tháng 7 2018
\(\frac{\frac{2}{3}-0,4-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-2,6-\frac{13}{7}+\frac{13}{11}}=\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
23 tháng 7 2018
\(\frac{\frac{2}{3}-0,4-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{2}{6}-\frac{13}{7}+\frac{13}{11}}=\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
Vậy \(\frac{\frac{2}{3}-0,4-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-2,6-\frac{13}{7}+\frac{13}{11}}=\frac{2}{13}\)
25 tháng 3 2017
a) = -3/7 . 5/11 + -3/7 . 6/11 + 9/7
= -3/7. ( 5/11 + 6/11 ) + 9/7
= -3/7. 1 + 9/7
= -3/7 + 9/7
= 6/7
b) = 4/13 + 9/13 + -11/5 + 6/5 - 3/4
= 13/13 + -5/5 - 3/4
= 1 + (-1) - 3/4
= 0 - 3/4
= -3/4
c) = -19/17. 4/7 + 19/17. -3/7 + 19/17
= 19/17. -4/7 + 19/17. -3/7 + 19/17.1
= 19/17.( -4/7 + -3/7 + 19/17
= 19/17. -7/7 + 19/17
= 19/17. (-1) + 19/17
= -19/17 + 19/17
= 0
tk mk nha,thanks
Đặt A=\(\frac{11}{5^2\cdot6^2}+\frac{13}{6^2\cdot7^2}+\frac{15}{7^2\cdot8^2}+...+\frac{39}{19^2\cdot20^2}\)
A=\(\frac{11}{25\cdot36}+\frac{13}{36\cdot49}+\frac{15}{49\cdot64}+...+\frac{39}{361\cdot400}\)
A=\(\frac{1}{25}-\frac{1}{36}+\frac{1}{36}-\frac{1}{49}+\frac{1}{49}-\frac{1}{64}+...+\frac{1}{361}-\frac{1}{400}\)
A=\(\frac{1}{25}-\frac{1}{400}\)
A=\(\frac{3}{80}\)
\(\Rightarrow\)A không phải là số nguyên
\(\frac{11}{5^2.6^2}+\frac{13}{6^2.7^2}+\frac{15}{7^2.8^2}+...+\frac{39}{19^2.20^2}\)
\(=\frac{11}{25.36}+\frac{13}{36.49}+\frac{15}{49.64}+...+\frac{39}{361.400}\)
\(=\frac{1}{25}-\frac{1}{36}+\frac{1}{36}-\frac{1}{49}+\frac{1}{49}-\frac{1}{64}+...+\frac{1}{361}-\frac{1}{400}\)
\(=\frac{1}{25}-\frac{1}{400}\)
\(=\frac{3}{80}\)
Mà \(\frac{3}{80}\notin Z\)
\(\Rightarrow\frac{11}{5^2+6^2}+\frac{13}{6^2.7^2}+\frac{15}{7^2.8^2}+...+\frac{39}{19^2.20^2}\notin Z\)