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\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)
\(\Leftrightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+5+2012}{2012}+\frac{x+4+2013}{2013}=\frac{x+3+2014}{2014}+\frac{x+2+2015}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)
\(\Rightarrow x+2017=0\)
\(\Rightarrow x=-2017\)
\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\cdot\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\)
Vì \(\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\ne0\)
suy ra \(x+2017=0\)
suy ra \(x=-2017\)
Vậy \(x=-2017\)
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 \(x=2015\)
có 2014/1+2013/2+2012/3+...+2/2013+1/2014=[1+(2013/2)]+[1+(2012/3)]+...+[1+(2/2013)]+[1+(1/2014)]+1
=2015/2+2015/3+...+2015/2014+2015/2015=2015.[1/2+1/3+..+1/2015)
vậy (1/2+1/3+...+1/2015).x=(1/2+1/3+...+1/2015).2015
x=2015
a) \(4+x=\frac{x+1}{5}\)
\(5.\left(4+x\right)=x+1\)
\(20+5.x=x+1\)
\(5.x-x=1-20\)
4.x = -19
x = -19/4
2) \(\frac{7}{x-1}=\frac{x}{8}\)
\(x.\left(x-1\right)=7.8\) ( x; x- 1 là 2 số tự nhiên liên tiếp)
=> x = 8
