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Ta co:1/n.1/n+1=1/n(n+1)=1/n^2+n;1/n-1/n+1=n+1/n(n+1)-n/n(n+1)=n+1-n/n^2+n=1/n^2+n
=>1/n.1/n+1=1/n-1/n+1
\(a)\)\(\frac{1}{n}\cdot\frac{1}{n+1}=\frac{1}{n(n+1)}\) ; \(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n(n+1)}=\frac{1}{n(n+1)}\)
\(b)A=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(A=\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}+\frac{1}{11\cdot12}\)
\(=(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{10})+(\frac{1}{10}-\frac{1}{11})+(\frac{1}{11}-\frac{1}{12})\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
a) Ta có hiệu của chúng là:
\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\left(1\right)\)
Mặt khác, ta lại có tích của chúng là:
\(\frac{1}{n}.\frac{1}{n+1}=\frac{1}{n\left(n+1\right)}\left(2\right)\)
Từ (1) và (2) suy ra: \(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}.\frac{1}{n+1}\)
Vậy tích của hai phân số này bằng hiệu của chúng (hiệu của phân số lớn trừ phân số nhỏ)
b) \(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}+\frac{1}{132}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}+\frac{1}{11.12}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(=\frac{1}{5}-\frac{1}{12}=\frac{7}{60}\)
a)\(\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n-1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
\(\Rightarrow\frac{1}{n\left(n+1\right)}=\frac{1}{n}.\frac{1}{n+1}\)
b) \(C=\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+\frac{1}{5}.\frac{1}{6}+\frac{1}{6}.\frac{1}{7}+\frac{1}{7}.\frac{1}{8}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{2}+0+0+0+0+0-\frac{1}{8}\)
\(=\frac{1}{2}-\frac{1}{8}=\frac{4}{8}-\frac{1}{8}=\frac{4-1}{8}=\frac{3}{8}\)
1.a.ta có:\(\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
mà \(\frac{2017}{2018}>\frac{2017}{2018+2019};\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow M>N\)
b.ta thấy:
\(\frac{n+1}{n+2}>\frac{n+1}{n+3}>\frac{n}{n+3}\Rightarrow\frac{n+1}{n+2}>\frac{n}{n+3}\)
=> A>B
\(1.a.\frac{x}{7}=\frac{6}{21}=\frac{6:3}{21:3}=\frac{2}{7}\Rightarrow x=2\\ b.\frac{-5}{y}=\frac{20}{28}=\frac{20:\left(-4\right)}{28:\left(-4\right)}=\frac{-5}{-7}\Rightarrow y=-7\)
\(2.a.\frac{a}{-b}=\frac{a\left(-1\right)}{-b\left(-1\right)}=\frac{-\left(a.1\right)}{-\left[-\left(b.1\right)\right]}=\frac{-a}{b}\\ b.\frac{-a}{-b}=\frac{-a\left(-1\right)}{-b\left(-1\right)}=\frac{-\left[-\left(a.1\right)\right]}{-\left[-\left(b.1\right)\right]}=\frac{a}{b}\)
\(3.\frac{3}{-4}=\frac{-3}{4}\\ \frac{-5}{-7}=\frac{5}{7}\\ \frac{2}{-9}=\frac{-2}{9}\\ \frac{-11}{-10}=\frac{11}{10}\)
\(4.\frac{3}{6}=\frac{2}{4}\\ \frac{6}{3}=\frac{4}{2}\\ \frac{2}{3}=\frac{4}{6}\\ \frac{3}{2}=\frac{6}{4}\)
Bài 1:
a, \(\frac{x}{7}\)=\(\frac{6}{21}\)⇒x.21=6.7⇒x.21=42⇒x=2
b,\(\frac{-5}{y}=\frac{20}{28}\)⇒-5.28= 20.y⇒-140=20.y⇒y =-7
Bài 2:
a, \(\frac{a}{-b}\)= \(\frac{a.\left(-1\right)}{-b.\left(-1\right)}\)=\(\frac{-a}{b}\)
b, \(\frac{-a}{-b}=\frac{-a.\left(-1\right)}{-b.\left(-1\right)}=\frac{a}{b}\)
Bài 3:
1,\(\frac{3}{-4}=\frac{-3}{4}\)
2,\(\frac{-5}{-7}=\frac{5}{7}\)
3,\(\frac{2}{-9}=\frac{-2}{9}\)
4,\(\frac{-11}{-10}=\frac{11}{10}\)
Bài 4 :
\(\frac{3}{6}=\frac{2}{4}\) ;
\(\frac{6}{3}=\frac{4}{2}\);
\(\frac{3}{2}=\frac{6}{4}\);
\(\frac{2}{3}=\frac{4}{6}\).
1/
+) \(\frac{3}{6}=\frac{2}{4};\frac{3}{2}=\frac{6}{4};\frac{4}{6}=\frac{2}{3};\frac{4}{2}=\frac{6}{3}\)
2/
\(A=\frac{3n-5}{n+4}=\frac{3n+12-17}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{17}{n+4}=3-\frac{17}{n+4}\)
Để A nguyên <=> n + 4 thuộc Ư(17) = {1;-1;17;-17}
| n+4 | 1 | -1 | 17 | -17 |
| n | -3 | -5 | 13 | -21 |
Vậy...
3/
\(S=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}\)
\(=1-\frac{1}{2017}\)
\(=\frac{2016}{2017}\)
\(A=\frac{3n+12-7}{n+4}=\frac{3\left(n+4\right)}{n+4}-\frac{7}{n+4}=3-\frac{7}{n+4}\)
=> n-4 \(\in\) Ư (7)
n-4=1
n=4+1=5
n-4=-1
n=-1+4=3
n-4=7
n=4+7=11
n-4=-7
n=-7+4=-3