Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\frac{x+6}{2010}+\frac{x+5}{2009}=\frac{x+4}{2008}+\frac{x+3}{2007}\)
\(\Leftrightarrow\frac{x+6}{2010}-1+\frac{x+5}{2009}-1=\frac{x+4}{2008}-1+\frac{x+3}{2007}-1\)
\(\Leftrightarrow\frac{x-2004}{2010}+\frac{x-2004}{2009}=\frac{x-2004}{2008}+\frac{x-2004}{2007}\)
\(\Leftrightarrow\frac{x-2004}{2010}+\frac{x-2004}{2009}-\frac{x-2004}{2008}-\frac{x-2004}{2007}=0\)
\(\Leftrightarrow\left(x-2004\right)\left(\frac{1}{2010}+\frac{1}{2009}-\frac{1}{2008}-\frac{1}{2007}\right)=0\)
Mà : \(\frac{1}{2010}+\frac{1}{2009}-\frac{1}{2008}-\frac{1}{2007}\ne0\)
Nên : x - 2004 = 0
=> x = 2004
Lời giải:
$\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}$
$\Leftrightarrow \frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1$
$\Leftrightarrow \frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}$
$\Leftrightarrow (x-2012)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0$
Dễ thấy $\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}< 0$
Do đó $x-2012=0\Rightarrow x=2012$
\(\frac{x-1}{2011}+\frac{x-2}{2010}-\frac{x-3}{2009}=\frac{x-4}{2008}.\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-4}{2008}+\frac{x-3}{2009}\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)=\left(\frac{x-4}{2008}-1\right)+\left(\frac{x-3}{2009}-1\right)\)
\(\Rightarrow\left(\frac{x-1-2011}{2011}\right)+\left(\frac{x-2-2010}{2010}\right)=\left(\frac{x-4-2008}{2008}\right)+\left(\frac{x-3-2009}{2009}\right)\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2008}+\frac{x-2012}{2009}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2008}-\frac{x-2012}{2009}=0\)
\(\Rightarrow\left(x-2012\right).\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2008}-\frac{1}{2009}\ne0.\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=0+2012\)
\(\Rightarrow x=2012\)
Vậy \(x=2012.\)
Chúc bạn học tốt!
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}+\frac{x+3}{2009}\)
=> \(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}-\frac{x+3}{2009}=0\)
=> \(\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
=> x + 3 = 0
=> x = 0 - 3
=> x = -3
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2010=0\)
\(\Leftrightarrow x=2010\)
\(\Rightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Rightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Rightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)
Vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\ne0\)
nên \(x-2010=0\Leftrightarrow x=2010\)
1) \(\frac{x+1}{15}+\frac{x+2}{14}=\frac{x+3}{13}+\frac{x+4}{12}\)
\(\Leftrightarrow\frac{x+16}{15}+\frac{x+16}{14}-\frac{x+16}{13}-\frac{x+16}{12}=0\)
\(\Leftrightarrow\left(x+16\right)\left(\frac{1}{15}+\frac{1}{14}-\frac{1}{13}-\frac{1}{12}\right)=0\)
\(\Leftrightarrow x=-16\)
2)3)4) tương tự
Gợi ý : 2) cộng 3 vào cả hai vế
3)4) cộng 2 vào cả hai vế
5) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=-3\)
\(\Leftrightarrow\frac{x+21}{20}+\frac{x+21}{19}+\frac{x+21}{18}=0\)
\(\Leftrightarrow\left(x+21\right)\left(\frac{1}{20}+\frac{1}{19}+\frac{1}{18}\right)=0\)
\(\Leftrightarrow x=-21\)
6) sửa VT = 4 rồi tương tự câu 5)
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Leftrightarrow\)\(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)
\(\Leftrightarrow\) \(\left(x+3\right).\left(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
\(\Leftrightarrow\) \(x+3=0\) ( Vì \(\frac{1}{2007}+\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\) )
\(\Leftrightarrow\) \(x=-3\)
Vậy x = -3
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Rightarrow\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)
\(\Rightarrow\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
\(\Rightarrow x+3=0\Leftrightarrow x=-3\)
\(\Rightarrow\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=0\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1+\frac{x-4}{2008}-1=0\)
\(\Rightarrow\left(\frac{x-1}{2011}-1\right)+\left(\frac{x-2}{2010}-1\right)+\left(\frac{x-3}{2009}-1\right)+\left(\frac{x-4}{2008}-1\right)=0\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}+\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\cdot\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)\)
Vì \(\frac{1}{2011}< \frac{1}{2009}\) và \(\frac{1}{2010}< \frac{1}{2008}\) nên \(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\ne0\)
\(\rightarrow x-2012=0\)
\(\rightarrow x=2012\)
Vậy x = 2012.
d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)
<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)
<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)
<=> x = -2010
\(\frac{x+3}{2007}-\frac{x+3}{2008}=\frac{x+3}{2010}-\frac{x+3}{2009}\)
\(\Leftrightarrow\)\(\frac{x+3}{2007}-\frac{x+3}{2008}-\frac{x+3}{2010}+\frac{x+3}{2009}=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2010}+\frac{1}{2009}\ne0\)
Nên \(x+3=0\)
\(\Leftrightarrow\)\(x=-3\)
Vậy \(x=-3\)
Chúc bạn học tốt ~