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\(\frac{x-2009-2010}{2008}+\frac{x-2008-2010}{2009}+\frac{x-2008-2009}{2010}=3\)
\(\Leftrightarrow\frac{x-2008-2009-2010}{2008}+\frac{x-2008-2009-2010}{2009}+\frac{x-2008-2009-2010}{2010}=0\)
\(\Leftrightarrow\left(x-2008-2009-2010\right)\left(\frac{1}{2008}-\frac{1}{2009}-\frac{1}{2010}\right)=0\)
\(\Leftrightarrow x-6027=0\Leftrightarrow x=6027\)
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)
<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)
<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0
<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0
<=> x+2011=0
<=> x=-2011
Vậy pt có nghiệm là x=-2011
\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)
\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)
\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)
\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)
\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)
3) Q=(3+1)(3^2+1)(3^4+1)....(3^3994+1)
=(3-1)(3+1)(3^2+1)(3^4+1)...(3^3994+1)
=(3^2-1)(3^2+1)(3^4+1)...(3^3994+1)
=(3^4-1)(3^4+1)...(3^3994+1)
=.........
=(3^3994-1)(3^3994+1)
=3^7988-1
áp dụng định lí Bê-du ta có:
R(x)=(-1)2009+(-1)2008+...+(-1)2+(-1)+2010=2009
xin lỗi tớ không biết kết quả tớ tính được có đúng không nhưng cách làm hình như đúng rồi đấy
Lời giải:
Xét PT(1):
\(\Leftrightarrow \frac{x-2013}{2011}+1+\frac{x-2011}{2009}+1=\frac{x-2009}{2007}+1+\frac{x-2007}{2005}+1\)
\(\Leftrightarrow \frac{x-2}{2011}+\frac{x-2}{2009}=\frac{x-2}{2007}+\frac{x-2}{2005}\)
\(\Leftrightarrow (x-2)\left(\frac{1}{2011}+\frac{1}{2009}-\frac{1}{2007}-\frac{1}{2005}\right)=0\)
Dễ thấy $\frac{1}{2011}+\frac{1}{2009}-\frac{1}{2007}-\frac{1}{2005}\neq 0$ nên $x-2=0$
$\Rightarrow x=2$Xét $(2)$:\(\Leftrightarrow \frac{(x-2)(x+m)}{x-1}=0\)
Để $(1);(2)$ là 2 PT tương đương thì $(2)$ chỉ có nghiệm $x=2$
Điều này xảy ra khi $x+m=x-1$ hoặc $x+m=x-2\Leftrightarrow m=-1$ hoặc $m=-2$
Akai Haruma Giáo viên, mk không hiểu tại sao lại có m=-1, m=-2 vào nữa, mk tưởng với mọi m chứ??
\(\frac{x}{2008}+\frac{x+1}{2009}+...+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+...+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+...+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)=0\)
Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)\ne0\)
Nên \(x-2008=0\)
\(\Leftrightarrow x=2008\)
Vậy : \(x=2008\)
\(\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}-5=0\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+\left(\frac{x+2}{2010}-1\right)+\left(\frac{x+3}{2011}-1\right)+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+\frac{x-2008}{2010}+\frac{x-2008}{2011}+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\ne0\)
\(\Rightarrow x-2008=0\)\(\Leftrightarrow x=2008\)
Vậy \(x=2008\)
Ta có: \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
=>\(\frac{2-x}{2007}=\frac{1-x}{2008}-\frac{x}{2009}+1\)
=>\(\frac{2-x}{2007}=\left(\frac{1-x}{2008}+1\right)-\frac{x}{2009}+1-1\)
=>\(\frac{2-x}{2007}+1=\frac{1-x+2008}{2008}+\left(1-\frac{x}{2009}\right)\)
=>\(\frac{2-x+2007}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)
=>\(\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)
=>\(\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)
=>\(\left(2009-x\right).\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)
=>2009-x=-
=>x=2009
Vậy tập nghiệm của phương trình S=2009