Tìm x biết:
a)(x-5)x+2014-(x-5)x+2015
b)\(\frac{x-1}{2014}+\frac{x-2}{2013}+\frac{x-3}{2012}=\frac{x-10}{2005}+\frac{x-11}{2004}+\frac{x-12}{2003}\)
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28 tháng 9 2020
Đề đúng phải là:
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)
Cộng mỗi phân thức thêm 1, quy đồng rồi chuyển sang 1 vế ta được:
\(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+2015}{2003}=0\)
\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà BT tích sau luôn nhỏ hơn 0
=> x+2015=0 => x = -2015
28 tháng 9 2020
\(\frac{x+1}{2014}+\frac{x+2}{2013}+\frac{x+3}{2012}=\frac{x+10}{2005}+\frac{x+11}{2004}+\frac{x+12}{2003}\)( như này đúng không ? :)) )
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+2}{2013}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+10}{2005}+1\right)+\left(\frac{x+11}{2004}+1\right)+\left(\frac{x+12}{2003}+1\right)\)
<=> \(\frac{x+1+2014}{2014}+\frac{x+2+2013}{2013}+\frac{x+3+2012}{2012}=\frac{x+10+2005}{2005}+\frac{x+11+2004}{2004}+\frac{x+12+2003}{2003}\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}=\frac{x+2015}{2005}+\frac{x+2015}{2004}+\frac{x+12}{2003}\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2013}+\frac{x+2015}{2012}-\frac{x+2015}{2005}-\frac{x+2015}{2004}-\frac{x+12}{2003}=0\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2014}+\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)
=> x + 2015 = 0
=> x = -2015
TL
3
22 tháng 3 2018
\(\frac{x+5}{2012}+\frac{x+4}{2013}=\frac{x+3}{2014}+\frac{x+2}{2015}\)
\(\Leftrightarrow\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+5+2012}{2012}+\frac{x+4+2013}{2013}=\frac{x+3+2014}{2014}+\frac{x+2+2015}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}=\frac{x+2017}{2014}+\frac{x+2017}{2015}\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}>0\)
\(\Rightarrow x+2017=0\)
\(\Rightarrow x=-2017\)
22 tháng 3 2018
\(\frac{x+5}{2012}+1+\frac{x+4}{2013}+1=\frac{x+3}{2014}+1+\frac{x+2}{2015}+1\)
\(\frac{x+2017}{2012}+\frac{x+2017}{2013}-\frac{x+2017}{2014}-\frac{x+2017}{2015}=0\)
\(\left(x+2017\right)\cdot\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)\)
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 \(x=-2017\)
16 tháng 2 2020
a, \(\frac{x+5}{x-1}=\frac{x+1}{x-3}-\frac{8}{x^2-4x+3}\)
= \(\frac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\frac{x^2-1}{\left(x-1\right)\left(x-3\right)}-\frac{8}{\left(x-1\right)\left(x-3\right)}\)
( x + 5)(x - 3) = \(x^2-1\) - 8
x\(^2\) -3x + 5x -15 = \(x^2-9\)
= > \(x^2-x^2\) +2x = 15 - 9
=> 2x = 6
=> x = 3
DT
0
NL
2
IM
10 tháng 9 2016
\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)
\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)
\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)
\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)
Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)
\(\Rightarrow x+2016=0\)
\(\Rightarrow x=-2016\)
21 tháng 4 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) (x-5)x+2015 - (x-5)x+2014 =0
(x-5)x+2014(x-5 -1) =0
+ x -5 =0 => x =5
+ x -6 =0 => x =6
Vậy x = 5 hoặc x =6