1,tìm x,y biết:
x-2011/12+x-2011/20+x-2011/30+x-2011/42+x-2011/56+x-2011/72=16/9
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IA
28 tháng 2 2020
\(\left(x-2011\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2011=8\)
\(x=2019\)
LT
28 tháng 2 2020
\(\frac{x-2011}{12}+\frac{x-2011}{20}+\frac{x-2011}{30}+\frac{x-2011}{42}+\frac{x-2011}{56}+\frac{x-2011}{72}=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2011\right)\frac{2}{9}=\frac{16}{9}\)
\(x-2011=8\Rightarrow x=2019\)
DT
1
N
20 tháng 12 2018
\(\Leftrightarrow2011.\left|x-2011\right|+\left(x-2011\right)^2=2013.\left|x-2011\right|\)
\(\Leftrightarrow\left(x-2011\right)^2=2.\left|x-2011\right|\Rightarrow\orbr{\begin{cases}x-2011=0\\2=\left|x-2011\right|\end{cases}}\Rightarrow\orbr{\begin{cases}x=2013\\x=2009\end{cases}}\text{hoặc }x=2011\)
Vậy ....
T
17 tháng 9 2018
a) \(\frac{x+4}{2009}+1+\frac{x+3}{2010}+1=\frac{x+2}{2011}+1+\frac{x+1}{2012}\)
\(\frac{x+4+2009}{2009}+\frac{x+3+2010}{2010}=\frac{x+2+2011}{2011}+\frac{x+2+2012}{2012}\)
\(\frac{x+2013}{2009}+\frac{x+2013}{2010}-\frac{x+2013}{2011}-\frac{x+2013}{2012}=0\)
\(\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\) (1)
Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)
Nên biểu thức (1) xảy ra khi \(x+2013=0\)
\(x=-2013\)
b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\) (2)
Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)
Nên biểu thức (2) xảy ra khi \(x-2011=0\)
\(x=2011\)
HG
1
PT
1
HG
2
IM
16 tháng 9 2016
\(\left(x-2011\right)^{x+1}-\left(x-2011\right)^{x+2011}=0\)
\(\Rightarrow\left(x-2011\right)^{x+1}.\left(1-x^{2010}\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-2011\right)^{x+1}=0\\1-x^{2010}=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=2011\\x=\pm1\end{array}\right.\)
Vậy x = 2011 ; x = 1 ; x = - 1
CN
4
NA
4 tháng 1 2017
Ta cÓ : ( x - 2011) x+ 1 - ( x - 2011)x + 2011
=) x - 2011= 0 =) x = 2011
DA
10 tháng 8 2017
X x 2011-X=2011x2009+2011
Xx(2011-1)=2011x(2009+1)
Xx2010 = 2011x2010
=> X=2011
