Rút gọn :
\(A=\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+....+\frac{403}{3^{100}}< \frac{17}{4}\)
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DY
1
20 tháng 4 2015
ta có: L = \(\frac{7}{3}+\frac{11}{3^2}+\frac{15}{3^3}+...+\frac{403}{3^{100}}\)
<=> \(3L=7+\frac{11}{3}+\frac{15}{3^2} +..+\frac{403}{3^{99}}\)
=> \(3L-L=\left(7+\frac{11}{3}+\frac{15}{3^2}+...+\frac{403}{3^{99}}\right)-\left(\frac{7}{3}+\frac{11}{3^2}+...+\frac{403}{3^{100}}\right)\)
<=> \(2L=7+\left(\frac{11}{3}-\frac{7}{3}\right)+\left(\frac{15}{3^2}-\frac{11}{3^2}\right)+...+\left(\frac{403}{3 ^{99}}-\frac{399}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=> \(2L=7+4\cdot\frac{1}{3}+4\cdot\frac{1}{3^2}+..+4\cdot\frac{1}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(2L=7+4\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\right)-\frac{403}{3^{100}}\)
<=>\(2L=7+4\left[\frac{1}{2}\cdot\left(1-\frac{1}{3^{99}}\right)\right]-\frac{403}{3^{100}}\)
<=> \(2L=7+2-\frac{2}{3^{99}}-\frac{403}{3^{100}}\)
<=> \(L=3,5+1-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}\)
<=> \(L=4,5-\frac{1}{3^{99}}-\frac{403}{2\cdot3^{100}}<4,5\)
1 ĐÚNG NHÉ
27 tháng 7 2019
a) \(A=\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A=4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{100}}\)
\(\Rightarrow3A-A=\left(4+\frac{7}{3}+\frac{10}{3^2}+...+\frac{301}{3^{99}}\right)-\left(\frac{4}{3}+\frac{7}{3^2}+...+\frac{301}{3^{100}}\right)\)
\(\Rightarrow2A=4+1+\frac{1}{3}+...+\frac{1}{3^{98}}-\frac{301}{3^{100}}\)
Đặt \(F=1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3F=3+1+...+\frac{1}{3^{97}}\)
\(\Rightarrow3F-F=\left(3+...+\frac{1}{3^{97}}\right)-\left(1+...+\frac{1}{3^{98}}\right)\)
\(\Rightarrow2F=3-\frac{1}{3^{98}}< 3\)
\(\Rightarrow F< \frac{3}{2}\)
\(\Rightarrow2A< 4+\frac{3}{2}\)
\(\Rightarrow2A< \frac{11}{2}\)
\(\Rightarrow A< \frac{11}{4}\left(đpcm\right)\)
27 tháng 7 2019
2. \(B=\frac{11}{3}+\frac{17}{3^2}+\frac{23}{3^3}+...+\frac{605}{3^{100}}\)
\(\Rightarrow3B=11+\frac{17}{3}+\frac{23}{3^2}+...+\frac{605}{3^{99}}\)
\(\Rightarrow3B-B=\left(11+...+\frac{605}{3^{99}}\right)-\left(\frac{11}{3}+...+\frac{605}{3^{100}}\right)\)
\(\Rightarrow2B=11+2+\frac{2}{3}+...+\frac{2}{3^{98}}-\frac{605}{3^{100}}\)
Đặt \(D=2+\frac{2}{3}+...+\frac{2}{3^{98}}\)
\(\Rightarrow3D=6+2+...+\frac{2}{3^{97}}\)
\(\Rightarrow2D=6-\frac{2}{3^{98}}< 6\)( làm tắt )
\(\Rightarrow2D< 6\)
\(\Rightarrow D< 3\)
\(\Rightarrow2B< 11+3\)
\(\Rightarrow2B< 14\)
\(\Rightarrow B< 7\left(đpcm\right)\)
15 tháng 3 2019
\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)
\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)
\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)
\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)
\(\Rightarrow-1\le x< 6\)
\(\Rightarrow x\in\left\{-1;0;1;2;3;4;5\right\}\)
Bài b tương tự
23 tháng 1 2019
\(A=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}\)
\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}\)
\(=\frac{2}{3}\)
16 tháng 5 2020
tìm n 
giúp mik với sắp thi r
LT
0
U
20 tháng 8 2018
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
20 tháng 8 2018
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
19 tháng 2 2017
1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
30 tháng 5 2015
a) \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)< \(x\) < \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
Ta có: \(\frac{-8}{3}+\frac{7}{5}+\frac{-71}{15}\)
=\(\frac{-40}{15}+\frac{21}{15}+\frac{-71}{15}\)
=\(\frac{-90}{15}\)
=\(-6\)
Ta có: \(\frac{-13}{7}+\frac{19}{14}+\frac{-7}{2}\)
=\(\frac{-26}{14}+\frac{19}{14}+\frac{-49}{14}\)
=\(\frac{-56}{14}\)
=\(-4\)
=> \(-6\)< \(x\)<\(-4\)
=> \(x=-5\)
b)\(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)< \(\frac{x}{9}\)<\(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
Ta có: \(\frac{5}{17}+\frac{-4}{9}+\frac{-20}{31}+\frac{12}{17}+\frac{-11}{31}\)
=\(\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}+\frac{-11}{31}\right)+\frac{-4}{9}\)
=\(1+\left(-1\right)+\frac{-4}{9}\)
=\(0+\frac{-4}{9}\)
=\(\frac{-4}{9}\)
Ta có: \(\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}+\frac{2}{3}\)
=\(\frac{-3}{7}+\frac{7}{15}+\frac{-4}{7}+\frac{8}{15}+\frac{2}{3}\)
=\(\left(\frac{-3}{7}+\frac{-4}{7}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{2}{3}\)
=\(\left(-1\right)+1+\frac{2}{3}\)
=\(0+\frac{2}{3}\)
=\(\frac{2}{3}\)
=> \(\frac{-4}{9}\)< \(\frac{x}{9}\)<\(\frac{2}{3}\)
=
=> \(\frac{-4}{9}\)<\(\frac{x}{9}\)<\(\frac{6}{9}\)
=> \(-4\)< \(x\)<\(6\)
=>\(x\in\left\{-3;-2;-1;0;1;2;3;4;5\right\}\)