Tìm x,y
a . (x+ 4/ 2007)+ (x+3/2008)=(x+2/2009)+(x+1/2010)
b . 2x+1 . 3y-2 = 12x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
OY
1
LT
14 tháng 3 2017
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(=>\dfrac{x+4}{2010}+1\))+(\(\dfrac{x+3}{2011}+1\))=\(\left(\dfrac{x+2}{2012}+1\right)\)+\(\left(\dfrac{x+1}{2013}+1\right)\)
=>\(\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
=>x+2014(\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\))=0
ta thấy \(\dfrac{1}{2010}>\dfrac{1}{2011}>\dfrac{1}{2012}>\dfrac{1}{2013}\)
=>\(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}>0\)
để A=0
\(\Leftrightarrow x+2014=0\)
\(\Leftrightarrow\)x=-2014
14 tháng 3 2017
a)\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\)\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)Mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
\(\Rightarrow x+2014=0\)
\(\Rightarrow x=-2014\)
NT
1
DN
1 tháng 8 2015
(x-4)/2007 + (x-3)/2008)= (x-2)/2009 + (x-1)/2010
=[(x-4)/2007 -1]+[(x-3)/2008 -1]=[(x-2)/2009 -1]+[(x-1)/2010 -1]
=(x-2011)/2007+(x-2011)/2008=(x-2011)/...
=(x-2011)(1/2007+1/2008-1/2009-1/2010)...
suy ra x=2010
TG
1 tháng 11 2016
ai lam guip toi cau nay voi mai toi nop bai roi
so sanh 2 phan so sau bang cach nahnh nhat: 2007/2008 voi 2008/2009
TN
12 tháng 11 2016
a)\(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1-\frac{x}{2009}+1\)
\(\Leftrightarrow\frac{2-x}{2007}+\frac{2007}{2007}=\frac{1-x}{2008}+\frac{2008}{2008}-\frac{x}{2009}+\frac{2009}{2009}\)
\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}-\frac{2009-x}{2009}\)
\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}+\frac{2009-x}{2009}=0\)
\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}+\frac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\).Do \(\frac{1}{2007}-\frac{1}{2008}+\frac{1}{2009}\ne0\)
\(\Leftrightarrow x=2009\)
b)\(\left(12x+7\right)^2\left(3x+2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow\left(12^2x^2+2\cdot12\cdot7x+7^2\right)\left(6x^2+7x+2\right)-3=0\)
\(\Leftrightarrow\left[24\left(6x^2+7x+2\right)+1\right]\left(6x^2+7x+2\right)-3=0\)
Đặt \(t=6x^2+7x+2\) ta có:
\(\left(24t+1\right)t-3=0\)\(\Leftrightarrow12t^2+t-3=0\)
Suy ra t rồi tìm đc x
27 tháng 4 2016
a) x+2x+3x+4x+...+2011x = 2012.2013
\(\Rightarrow\) x(1+2+3+4+...+2011) = 4050156
\(\Rightarrow\) x.2023066 = 4050156
\(\Rightarrow\) x = 4026/2011
a) Theo đề bài, ta có:
\(\dfrac{x+4}{2007}+\dfrac{x+3}{2008}=\dfrac{x+2}{2009}+\dfrac{x+1}{2010}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2007}+1\right)+\left(\dfrac{x+3}{2008}+1\right)=\left(\dfrac{x+2}{2009}+1\right)+\left(\dfrac{x+1}{2010}+1\right)\)
\(\Leftrightarrow\left(\dfrac{x+2011}{2007}+\dfrac{x+2011}{2008}\right)-\left(\dfrac{x+2011}{2009}+\dfrac{x+2011}{2010}\right)=0\)
\(\Leftrightarrow\left(x+2011\right)\left(\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\right)=0\)
\(\Leftrightarrow x+2011=0\)
\(\) Vậy \(x=-2011\)