2+\(\frac{2}{3}\)+\(\frac{2}{6}\)+\(\frac{2}{12}\)+.....+\(\frac{2}{x\left(x+1\right)}\)=1\(\frac{19}{19}\)
K
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14 tháng 9 2016
b) \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=\frac{-2}{3}\)
d) \(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}+\frac{2}{11}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}+\frac{13}{11}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{11}\right)}=\frac{2}{13}\)
15 tháng 9 2016
Làm tiếp:
\(=\left(1+\frac{1}{2}+.....+\frac{1}{2017}\right)-\left(1+\frac{1}{2}+....+\frac{1}{1008}\right)\)
\(=\frac{1}{1009}+\frac{1}{1010}+.........+\frac{1}{2017}\)
\(\Rightarrow\frac{\frac{1}{1009}+....+\frac{1}{2017}}{1-\frac{1}{2}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}}=1\)
Bài 2:
Đặt \(A=\frac{1}{2^2}+.......+\frac{1}{2^{800}}\)
\(4A=1+\frac{1}{2^2}+.....+\frac{1}{2^{798}}\)
\(\Rightarrow4A-A=1-\frac{1}{2^{800}}\)
\(\Rightarrow3A=1-\frac{1}{2^{800}}< 1\Rightarrow A< \frac{1}{3}\)
Vậy \(\frac{1}{2^2}+\frac{1}{2^4}+........+\frac{1}{2^{800}}< \frac{1}{3}\)
15 tháng 9 2016
Bài 1:Tính
a, Xét biểu thức \(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).........\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)..........\left(1+\frac{n+2}{n}\right)}\) với\(n\in N\)
Ta có:\(\frac{\left(1+\frac{n}{1}\right)\left(1+\frac{n}{2}\right).......\left(1+\frac{n}{n+2}\right)}{\left(1+\frac{n+2}{1}\right)\left(1+\frac{n+2}{2}\right)......\left(1+\frac{n+2}{n}\right)}\)
\(=\frac{\frac{n+1}{1}.\frac{n+2}{2}........\frac{2n+2}{n+2}}{\frac{n+3}{1}.\frac{n+4}{2}.........\frac{2n+2}{n}}\)
\(=\frac{\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right)}{1.2.3.........\left(n+2\right)}}{\frac{\left(n+3\right)\left(n+4\right)........\left(2n+2\right)}{1.2.3.........n}}\)
\(=\frac{\left(n+1\right)\left(n+2\right).......\left(2n+2\right).1.2.3.......n}{\left(n+3\right)\left(n+4\right)........\left(2n+2\right).1.2.3......\left(n+2\right)}\)
\(=\frac{\left(n+1\right)\left(n+2\right)}{\left(n+1\right)\left(n+2\right)}=1\)
Áp dụng vào bài toán ta có đáp số là:1
b, \(\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}=\frac{\left(-6\right).\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}=\frac{-6}{9}=-\frac{2}{3}\)
c,\(\frac{\frac{1}{6}-\frac{1}{39}+\frac{1}{51}}{\frac{1}{8}-\frac{1}{52}+\frac{1}{68}}=\frac{\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}{\frac{1}{4}.\left(\frac{1}{2}-\frac{1}{13}+\frac{1}{17}\right)}=\frac{\frac{1}{3}}{\frac{1}{4}}=12\)
d,\(\frac{\frac{2}{3}-\frac{2}{5}-\frac{2}{7}}{\frac{13}{3}-\frac{13}{5}-\frac{13}{7}}=\frac{2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}{13\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}\right)}=\frac{2}{13}\)
e,Xét mẫu số ta có:
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+..........+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}\)
\(=1+\frac{1}{2}-2.\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-2.\frac{1}{4}+.....+\frac{1}{2015}+\frac{1}{2016}-2.\frac{1}{2016}+\frac{1}{2017}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.......+\frac{1}{2017}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+.........+\frac{1}{2016}\right)\)
KN
3 tháng 8 2020
\(-\frac{5}{6}x-1\frac{13}{19}:3\frac{7}{19}-\left(-0,12\right)=\left(-\frac{1}{2}\right)^2\)
\(\Rightarrow-\frac{5}{6}x-\frac{32}{19}:\frac{64}{19}+0,12=\frac{1}{4}\)
\(\Rightarrow-\frac{5}{6}x-\frac{1}{2}=\frac{1}{4}-0,12=\frac{13}{100}\)
\(\Rightarrow-\frac{5}{6}x=\frac{63}{100}\)
\(\Rightarrow x=-\frac{189}{250}\)
3 tháng 8 2020
-5/6x-32/19:64/19+3/25=1/4
-5/6x-32/19×19/64+3/25=1/4
-5/6x-1/2+3/25=1/4
-5/6x-31/50=1/4
-5/6x=1/4+31/50
-5/6x=87/100
x=87/100:-5/6
x=87/100×-6/5
x=-261/250
vậy x=-261/250
Ê hỗn số \(1\frac{19}{19}\)là sai vì 19 vẫn chia típ đc cho 19
\(2+\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=2\)
\(\Rightarrow2\left(1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=2-\frac{2}{3}\)
\(\Rightarrow2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}+\frac{1}{x+1}\right)=\frac{4}{3}\)
\(\Rightarrow2\left(1+\frac{1}{2}-\frac{1}{x+1}\right)=\frac{4}{3}\)
\(\Rightarrow3+\frac{1}{x+1}=\frac{4}{3}\)
\(\Rightarrow\frac{1}{x+1}=\frac{4}{3}-3\)
\(\Rightarrow\frac{1}{x+1}=\frac{-5}{3}\)
\(\Rightarrow x=\frac{-8}{5}\)
Mà \(x\in N\Rightarrow\)đề sai :D