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M lớn hơn hay nhỏ hơn N vậy bạn ơi??
Nếu m > n thì A > B; m < n thì A < B nhé!!
Bài 1 :
Ta có :
\(A=\frac{10^{17}+1}{10^{18}+1}=\frac{\left(10^{17}+1\right).10}{\left(10^{18}+1\right).10}=\frac{10^{18}+10}{10^{19}+10}\)
Mà : \(\frac{10^{18}+10}{10^{19}+10}>\frac{10^{18}+1}{10^{19}+1}\)
Mà \(A=\frac{10^{18}+10}{10^{19}+10}\)nên \(A>B\)
Vậy \(A>B\)
Bài 2 :
Ta có :
\(S=\frac{2013}{2014}+\frac{2014}{2015}+\frac{2015}{2016}+\frac{2016}{2013}\)
\(\Rightarrow S=\frac{2014-1}{2014}+\frac{2015-1}{2015}+\frac{2016-1}{2016}+\frac{2013+3}{2013}\)
\(\Rightarrow S=1-\frac{1}{2014}+1-\frac{1}{2015}+1-\frac{1}{2016}+1+\frac{3}{2013}\)
\(\Rightarrow S=4+\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)\)
Vì \(\frac{1}{2013}>\frac{1}{2014}>\frac{1}{2015}>\frac{1}{2016}\)nên \(\frac{3}{2013}-\left(\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)>0\)
Nên : \(M>4\)
Vậy \(M>4\)
Bài 3 :
Ta có :
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+.......+\frac{1}{100^2}\)
Suy ra : \(A< \frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+....+\frac{1}{99.101}\)
\(\Rightarrow A< \frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+......+\frac{2}{99.101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-......-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{99}\right)-\left(\frac{1}{3}+\frac{1}{4}+......+\frac{1}{101}\right)\right]\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)
\(\Rightarrow A< \frac{1}{2}.\left(1+\frac{1}{2}\right)\)
\(\Rightarrow A< \frac{3}{4}\)
Vậy \(A< \frac{3}{4}\)
Bài 4 :
\(a)A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2015.2017}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{1}{2015.2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{2015}-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{2017}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{2016}{2017}\)
\(\Rightarrow A=\frac{1008}{2017}\)
Vậy \(A=\frac{1008}{2017}\)
\(b)\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+......+\frac{1}{x\left(x+2\right)}=\frac{1008}{2017}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{x.\left(x+2\right)}=\frac{2016}{2017}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2016}{2017}\)
\(1-\frac{1}{x+2}=\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{2017}\)
\(\Rightarrow x+2=2017\)
\(\Rightarrow x=2017-2=2015\)
Vậy \(x=2015\)
a. Ta có
\(B=\frac{2011+2012}{2012+2013}=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}.\)
Vì\(\frac{2011}{2012+2013}< \frac{2011}{2012}.\)(1)
\(\frac{2012}{2012+2013}< \frac{2012}{2013}.\)(2)
Cộng vế với vế của 1;2 ta được
\(B=\frac{2011}{2012+2013}+\frac{2012}{2012+2013}< A=\frac{2011}{2012}+\frac{2012}{2013}\)
hay A>B
bài 3.\(\frac{x-7}{3}=\frac{4x-1}{2}\Leftrightarrow2x-14=12x-3\\ \Leftrightarrow10x=-11\\ \Leftrightarrow x=\frac{-11}{10}\)
Bài 1 :
ta thấy :
\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};......;\frac{1}{\left(n-1\right)^2}< \frac{1}{\left(n-2\right).\left(n-1\right)};\frac{1}{n^2}< \frac{1}{\left(n-1\right).n}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{\left(n-1\right)^2}+\frac{1}{n^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{\left(n-2\right).\left(n-1\right)}+\frac{1}{\left(n-1\right).n}\)
mà :
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{\left(n-2\right).\left(n-1\right)}+\frac{1}{\left(n-1\right).n}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\)
=\(1-\frac{1}{n}\)<1
=>A<1
Bài 3 :
\(\frac{x-7}{3}=\frac{4x-1}{2}\)
=>\(\frac{2.\left(x-7\right)}{6}=\frac{3\left(4x-1\right)}{6}\)
=>\(2\left(x-7\right)=3\left(4x-1\right)\)
=>\(x-7=\frac{3}{2}.\left(4x-1\right)\)
=>\(\frac{x-7}{4x-1}=\frac{3}{2}\)
\(=>\left\{{}\begin{matrix}x-7=3=>x=10\\4x-1=2=>4x=3=>x=\frac{3}{4}\end{matrix}\right.\)
Vậy x ∈{\(10;\frac{3}{4}\)}
B,
(1 - x-1/2011)+(1 - x-2/2012)+(1 - x-3/2013)=(1 - x-4/2014)+(1 - x-5/2015)+(1 - x-6/2016)
=> 2010-x/2011 + 2010-x/2012 + 2010-x/2013 = 2010-x/2014 + 2010-x/2015 + 2010-x/2016
=> 2010-x/2011 + 2010-x/2012 + 2010-x/2013 - 2010-x/2014 - 2010-x/2015 - 2010-x/2016=0
=>(2010-x).(1/2011 + 1/2012 + 1/2013 + 1/2014 - 1/2015 - 1/2016)=0
Mà: 1/2011 + 1/2012 + 1/2013 + 1/2014 - 1/2015 - 1/2016 khác 0
=> 2010-x=0
=>x=2010
a, 10/a^m > 11/a^m; 10/a^n > 9/a^n => A > B
b, bạn cộng 1 vào các phân số đưa VP qua VT đặt nhân tử chung x + 2010 thì trong ngoặc còn lại là số dương nên x + 2010 = 0