Cho biểu thức
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
chứng minh rằng \(M< \frac{1}{4}\)
K
Khách
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22 tháng 10 2016
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)
\(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\)
\(5M=5\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(5M=1+\frac{1}{5}+...+\frac{1}{5^{49}}\)
\(5M-M=\left(1+\frac{1}{5}+...+\frac{1}{5^{49}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{50}}\right)\)
\(4M=1-\frac{1}{5^{50}}\)
\(M=\frac{1-\frac{1}{5^{50}}}{4}< \frac{1}{4}=0,25\)
Đpcm
8 tháng 11 2016
Bài 4:
Giải:
Vì Om là tia phân giác của góc xOz nên:
mOz = 1/2.xOz
Vì On là tia phân giác của góc zOy nên:
zOn = 1/2 . zOy
Ta có: xOz + zOy = 180o ( kề bù )
=> 1/2(xOz + zOy) = 1/2 . 180o
=> 1/2.xOz + 1/2.zOy = 90o
=> mOz + zOn = 90o
=> mOn = 90o (đpcm)
8 tháng 11 2016
Bài 2:
7^6 + 7^5 - 7^4 = 7^4.( 7^2 + 7 - 1 ) = 7^4 . 55 chia hết cho 55
Vậy 7^6 + 7^5 - 7^4 chia hết cho 55
A = 1 + 5 + 5^2 + ... + 5^50
=> 5A = 5 + 5^2 + 5^3 + ... + 5^51
=> 5A - A = ( 5 + 5^2 + 5^3 + ... + 5^51 ) - ( 1 + 5 + 5^2 + ... + 5^50 )
=> 4A = 5^51 - 1
=> A = ( 5^51 - 1 )/4
22 tháng 3 2020
\(A=\left[\frac{1\frac{11}{31}\cdot4\frac{3}{7}-\left(15-6\frac{1}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-5\frac{1}{3}\right)}\cdot\left(-1\frac{14}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{\frac{42}{31}\cdot\frac{31}{7}-\left(15-\frac{19}{3}\cdot\frac{2}{19}\right)}{4\frac{5}{6}+\frac{1}{6}\left(12-\frac{16}{3}\right)}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\left(15-\frac{2}{3}\right)}{\frac{29}{6}+\frac{10}{9}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{6-\frac{43}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\left[\frac{-\frac{25}{3}}{\frac{107}{18}}\cdot\left(-\frac{107}{93}\right)\right]\cdot\frac{31}{50}\)
\(A=\frac{50}{31}\cdot\frac{31}{50}=1\)
KN
2 tháng 11 2019
a) Ta có:
\(\frac{1}{2\left(m+1\right)}+\frac{1}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3m+2}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(m+1\right)\left(3m+2\right)}\)
\(+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3m+3}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3\left(m+1\right)}{2\left(m+1\right)\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3}{2\left(3m+2\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{3\left(8m+5\right)}{2\left(3m+2\right)\left(8m+5\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{24m+15}{2\left(3m+2\right)\left(8m+5\right)}+\frac{1}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{24m+16}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{8\left(3m+2\right)}{2\left(3m+2\right)\left(8m+5\right)}\)
\(=\frac{8}{2\left(8m+5\right)}=\frac{4}{8m+5}\left(đpcm\right)\)
KN
2 tháng 11 2019
b) Ta có: \(\frac{1}{m+1}+\frac{1}{3m+2}+\frac{1}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{3m+2}{\left(m+1\right)\left(3m+2\right)}+\frac{m+1}{\left(m+1\right)\left(3m+2\right)}\)
\(+\frac{1}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{4m+4}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{4\left(m+1\right)}{\left(m+1\right)\left(3m+2\right)}\)
\(=\frac{4}{3m+2}\left(đpcm\right)\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
\(\begin{array}{l}A = \left( {7 - \frac{2}{5} + \frac{1}{3}} \right) - \left( {6 - \frac{4}{3} + \frac{6}{5}} \right) - \left( {2 - \frac{8}{5} + \frac{5}{3}} \right)\\A = 7 - \frac{2}{5} + \frac{1}{3} - 6 + \frac{4}{3} - \frac{6}{5} - 2 + \frac{8}{5} - \frac{5}{3}\\A = \left( {7 - 6 - 2} \right) + \left( { - \frac{2}{5} - \frac{6}{5} + \frac{8}{5}} \right) + \left( {\frac{1}{3} + \frac{4}{3} - \frac{5}{3}} \right)\\A = - 1 + 0 + 0 = - 1\end{array}\)
Chú ý:
Trong phép tính chỉ có phép cộng trừ, ta có thể đổi chỗ các số hạng tùy ý kèm theo dấu của chúng.
IA
25 tháng 3 2020
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok tốt
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^3+...+\left(\frac{1}{5}\right)^{49}+\left(\frac{1}{5}\right)^{50}\)
\(5M=1+\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{48}+\left(\frac{1}{5}\right)^{49}\)
5M - M = \(1-\left(\frac{1}{5}\right)^{50}\)hay 4M = \(1-\left(\frac{1}{5}\right)^{50}\)< 1
\(\Rightarrow M=\frac{1-\left(\frac{1}{5}\right)^{50}}{4}< \frac{1}{4}\)
\(M=\frac{1}{5}+\left(\frac{1}{5}\right)^2+...+\left(\frac{1}{5}\right)^{50}\)(1)
\(\Rightarrow5M=1+\frac{1}{5}+...+\left(\frac{1}{5}\right)^{49}\)(2)
Lấy (2)-(1) ta có
\(\Rightarrow4M=1-\left(\frac{1}{5}\right)^{50}\)
\(\Rightarrow M=\frac{1-\frac{1}{5^{50}}}{4}\)
Do \(1-\frac{1}{5^{50}}< 1\)
\(\Rightarrow M< \frac{1}{4}\)