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\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)\)
\(\Rightarrow\left(\frac{x+1}{2009}+\frac{2009}{2009}\right)+\left(\frac{x+2}{2008}+\frac{2008}{2008}\right)=\left(\frac{x+3}{2007}+\frac{2007}{2007}\right)+\left(\frac{x+4}{2006}\frac{2006}{2006}\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2007}+\frac{x+2010}{2006}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2007}-\frac{x+2010}{2006}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
=>x+2010=0
=>x=-2010
Vậy x = -2010
Trừ 1 đi ở mỗi phân số, ta có:
\(\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Rightarrow\frac{x-1}{2009}-\frac{2009}{2009}+\frac{x-2}{2008}-\frac{2008}{2008}=\frac{x-3}{2007}-\frac{2007}{2007}+\frac{x-4}{2006}-\frac{2006}{2006}\)
\(\Rightarrow\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}=\frac{x-3-2007}{2007}+\frac{x-4-2006}{2006}\)
\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Rightarrow\left[x-2010\right]\left[\frac{1}{2009}+\frac{1}{2008}\right]=\left[x-2010\right]\left[\frac{1}{2007}+\frac{1}{2006}\right]\)
Sẽ có hai trường hợp
TH1: Cả hai vế đều bằng 0
Ta có: \(\hept{\begin{cases}\frac{1}{2009}+\frac{1}{2008}\ne0\\\frac{1}{2007}+\frac{1}{2006}\ne0\end{cases}}\Rightarrow x-2010=0\Rightarrow x=2010\)
TH2: Cả hai vế khác 0
Ta bỏ đi x - 2010 vì cả hai bên đều có
\(\Rightarrow\frac{1}{2009}+\frac{1}{2008}=\frac{1}{2007}+\frac{1}{2006}\)Vô lí
Vậy x = 2010
X-1/2009 + X-2/2008 = X-3/2007 + X-4/2006
thôi nói cho nhanh nhé
bạn trừ 1 vào tất cả các giá trị VD: (X-1/2009)-1. Ta được tử chung là X-2010 cứ thế mà đặt ra làm thôi. Ko dc thì bảo tớ chỉ tiếp.
\(\Rightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Rightarrow\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}=\frac{x-3-2007}{2007}+\frac{x-4-2006}{2006}\)
\(\frac{x-1-2009}{2009}+\frac{x-2-2008}{2008}-\frac{x-3-2007}{2007}-\frac{x-4-2006}{2006}=0\)
\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
=>(x-2010)(1/2009+1/2008-1/2007-1/2006)=0
mà 1/2009+1/2008-1/2007-1/2006 khác 0
=>x-2010=0=>x=2010
cho mìh đi rồi gửi lại đề bài qua tin nhắn cho mìh, mìh sẽ giải cho bn
\(\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
Vì \(\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)\ne0\)
=> x-2010 =0
=> x =2010
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
a) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
<=> \(\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)-\left(\frac{x-3}{2007}-1\right)-\left(\frac{x-4}{2006}-1\right)=0\)
<=> \(\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
<=> \(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
<=> x - 2010 = 0 Vì \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\)
<=> x = 2010
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=2010\)
\(\Rightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Rightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\ne0\)
nên \(x-2010=0\Leftrightarrow x=2010\)
\(\left(\frac{x+4}{2007}+1\right)+\left(\frac{x+3}{2008}+1\right)=\left(\frac{x+2}{2009}+1\right)+\left(\frac{x+1}{2010}+1\right)\)
\(\left(\frac{x+2011}{2007}\right)+\left(\frac{x+2011}{2008}\right)=\left(\frac{x+2011}{2009}\right)+\left(\frac{x+2011}{2010}\right)\)
\(\frac{x+2011}{2007}+\frac{x+2011}{2008}-\frac{x+2011}{2009}-\frac{x+2011}{2010}=0\)
\(\left(x+2011\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\)khác 0 (các số hạng ko bằng nhau)
\(\Leftrightarrow\)\(x+2011=0\)
\(\Rightarrow x=0-2011\)
\(\Rightarrow x=-2011\)
Ta có:
x-1/2009 + x-2/2008 = x-3/2007 + x-4/2006
=> (x-1/2009 - 1)+(x-2/2008 - 1) = (x-3/2007 - 1) + (x-4/2006 -1 )
=> x-2010/2009 + x-2010/2008 = x-2010/2007 + x-2010/2006
=> x-2010/2009 + x-2010/2008 - (x-2010/2007 + x-2010/2006)=0
=> (x-2010)(1/2009+1/2008-1/2007-1/2006)=0
Vì (1/2009+1/2008-1/2007-1/2006) KHÁC 0
=>x-2010=0
=>x=2010
Cảm ơn bạn nhiều !