Tìm x , biết : \(\frac{x-2}{2014}+\frac{x-3}{2013}+\frac{x-4}{2012}=\frac{3x-15}{2011}\)
K
Khách
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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 \(x-2011=0\)
\(x=2011\)
9 tháng 10 2016
\(\frac{x+4}{2011}+\frac{x+3}{2012}=\frac{x+2}{2013}+\frac{x+1}{2014}\)
\(\Leftrightarrow\left(\frac{x+4}{2011}+1\right)+\left(\frac{x+3}{2012}+1\right)-\left(\frac{x+2}{2013}+1\right)-\left(\frac{x+1}{2014}+1\right)=0\)
\(\Leftrightarrow\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\Leftrightarrow x+2015=0\) (Vì: \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\) )
\(\Leftrightarrow x=-2015\)
NT
3
9 tháng 7 2015
\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)
\(\frac{x+1}{2014}+1+\frac{x+2}{2013}+1=\frac{x+3}{2012}+1+\frac{x+4}{2011}+1\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)
\(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\) nên x + 2015 = 0
x = 0 - 2015
x = -2015
9 tháng 7 2015
\(\frac{x+1}{2014}+\frac{x+2}{2013}=\frac{x+3}{2012}+\frac{x+4}{2011}\)
\(1+\frac{x+1}{2014}+1+\frac{x+2}{2013}=1+\frac{x+3}{2012}+1+\frac{x+4}{2011}\)
\(\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}=\frac{x+3+2012}{2012}+\frac{x+4+2011}{2011}\)
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)
\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)
\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)
=> x + 2015 = 0 ( vì 1/2014 + 1/2013 - 1/2012 - 1/2011 khác 0)
=> x = -2015
DL
25 tháng 6 2018
\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)
\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)
\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)
Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức
25 tháng 6 2018
\(a)\) \(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)
\(\Leftrightarrow\)\(\frac{x-1+2011}{2011}+\frac{x-2+2012}{2012}=\frac{x-3+2013}{2013}+\frac{x-4+2014}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)
\(\Leftrightarrow\)\(\frac{x-2010}{2011}+\frac{x+2010}{2012}-\frac{x+2010}{2013}-\frac{x+2010}{2014}=0\)
\(\Leftrightarrow\)\(\left(x-2010\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)
Nên \(x-2010=0\)
\(\Rightarrow\)\(x=2010\)
Vậy \(x=2010\)
Chúc bạn học tốt ~
VT
Tìm x, biết:
\(\frac{x+1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}=\frac{x-1}{2014}+\frac{x-1}{2015}\)
1
9 tháng 8 2018
\(\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}=\frac{x-1}{2014}+\frac{x-1}{2015}\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}-\frac{x-1}{2014}-\frac{x-1}{2015}=0\)
\(\left(x-1\right).\left(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
mà \(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
=> x - 1 = 0
x = 1
bn có chép sai đề ko z???
M
23 tháng 1 2016
a) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-1}{17}-5=0\)
\(\Leftrightarrow\frac{x-90-10}{10}+\frac{x-76-2.12}{12}+\frac{x-58-3.14}{14}+\frac{x-36-4.16}{16}+\frac{x-15-5.17}{17}=0\)
\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
b) \(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)
\(\Leftrightarrow\frac{x+2011}{2013}+1+\frac{x+2012}{2012}+1=\frac{x+2010}{2014}+1+\frac{x+2013}{2011}+1\)
\(\Leftrightarrow\frac{x+2011+2013}{2013}+\frac{x+2012+2012}{2012}=\frac{x+2010+2014}{2014}+\frac{x+2013+2011}{2011}\)
\(\Leftrightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)
\(\Leftrightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)
\(\Leftrightarrow x+4024=0\Leftrightarrow x=-4024\)
Vậy \(S=\left\{-4024\right\}\)
23 tháng 1 2016
Phương trình a bạn trừ phân thức đầu tiên cho 1, phân thức thứ hai cho 2, phân thức thứ ba cho 3, phân thức thứ tư cho 4, phân thức thứ năm cho 5, vế còn lại trừ đi 15. Tiếp theo bạn đặt x -100 làm nhân tử chung. Cuối cùng tìm được x= 100
LN
1
MT
18 tháng 6 2015
\(\frac{x+4}{2011}+\frac{x+3}{2012}=\frac{x+2}{2013}+\frac{x+1}{2014}\)
\(\frac{x+4}{2011}+1+\frac{x+3}{2012}+1=\frac{x+2}{2013}+1+\frac{x+1}{2014}+1\)
\(\frac{x+4}{2011}+\frac{2011}{2011}+\frac{x+3}{2012}+\frac{2012}{2012}=\frac{x+2}{2013}+\frac{2013}{2013}+\frac{x+1}{2014}+\frac{2014}{2014}\)
\(\frac{x+2015}{2011}+\frac{x+2015}{2012}=\frac{x+2015}{2013}+\frac{x+2015}{2014}\)
\(\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)
\(\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)
vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)nên:
x+2015=0
x=-2015
