Ta có : M = \(\frac{x+y}{z}+\frac{x+z}{y}=\frac{y+z}{x}\)
\(\Rightarrow M+3=\left(\frac{x+y}{z}+1\right)+\left(\frac{x+z}{y}+1\right)+\left(\frac{y+z}{x}+1\right)\)
\(\Rightarrow M+3=\frac{x+y+z}{z}+\frac{x+y+z}{y}+\frac{x+y+z}{x}\)
\(\Rightarrow M+3=\left(x+y+z\right).\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow M+3=2020.\frac{1}{202}\)
=> M + 3 = 10
=> M = 7
Vậy M = 7
b) Ta có : \(A=\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2017^2}\)
\(=\frac{2}{3.3}+\frac{2}{5.5}+\frac{2}{7.7}+...+\frac{2}{2017.2017}\)
\(< \frac{2}{\left(3+1\right)\left(3-1\right)}+\frac{2}{\left(5-1\right)\left(5+1\right)}+\frac{2}{\left(7-1\right)\left(7+1\right)}+...+\frac{2}{\left(2017-1\right)\left(2016-1\right)}\)
\(=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2016.2018}\)
\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2016}-\frac{1}{2018}\)
\(=\frac{1}{2}-\frac{1}{2018}\)
\(=\frac{1008}{2018}=\frac{504}{1009}\)
=> \(A< \frac{504}{1009}\left(\text{ĐPCM}\right)\)
Bài a ) \(\frac{2}{\left(3+1\right)\left(3-1\right)}\) là như thế nào bạn
Thế này cho dễ hiểu nha :
\(\frac{2}{3^2}=\frac{2}{9}< \frac{2}{9-1}=\frac{2}{8}=\frac{2}{2.4}\)
tương tự các số khác cũng như vậy
Không dậy mình hàng thứ 2 đó dễ hiểu hơn
Ta có : x + y + z = 2020
=> x+y = 2020- z
x + z = 2020 - y
y + z = 2020 - x
Khi đó M = \(\frac{2020-z}{z}+\frac{2020-y}{y}+\frac{2020-x}{x}=\frac{2020}{z}-1+\frac{2020}{y}-1+\frac{2020}{x}-1\)
\(=\left(\frac{2020}{x}+\frac{2020}{y}+\frac{2020}{z}\right)-3\)
\(=2020\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-3\)
\(=\frac{2020}{202}-3=10-3=7\)
=> M = 7
