câu a với câu b dễ mà , có thể câu b cần lập bảng xét dấu , dựa vào đó rồi suy ra các trường hợp cầm tìm , còn câu c bạn trừ cả 2 về cho 2 ta suy ra (x-2010)(1/2009 + 1/2008 -1/2007-1/2006) = 0
từ đó suy ra x=2010
có j khó hiểu có thể ib hỏi mik ........sorry ko giải kĩ ra được vì bây h mik đag bận !
a) Ta có:
\(\left(2x+1\right)^4=\left(2x+1\right)^6\)
\(\Leftrightarrow\left(2x+1\right)^4-\left(2x+1\right)^6=0\)
\(\Leftrightarrow\left(2x+1\right)^4\left[\left(2x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+1\right)^4=0\\\left(2x+1\right)^2-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-0,5\\\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-0,5\\x=0\\x=-1\end{matrix}\right.\)
b) Ta có:
\(|\left|x+3\right|-8|=20\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+3\right|-8=20\\\left|x+3\right|-8=-20\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}\left|x+3\right|=28\\\left|x+3\right|=-12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+3=28\\x+3=-28\end{matrix}\right.\\x+3\in\varnothing\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=25\\x=-31\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=25\\x=-31\end{matrix}\right.\)
c) \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Mà \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
\(\Rightarrow x-2010=0\Leftrightarrow x=2010\)
Vậy \(x=2010\)
a) (2x + 1)4 = (2x + 1)6
Mà 2 vế có chung 2x + 1
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\2x+1=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0-1=-1\\2x=1-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=0\end{matrix}\right.\)
b) ||x + 3| - 8| = 20
\(\Rightarrow\left[{}\begin{matrix}\left|x+3\right|-8=20\\\left|x+3\right|-8=-20\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x+3\right|=20+8=28\\\left|x+3\right|=-20+8=-12\end{matrix}\right.\)
Vì \(\left|x+3\right|\ge0\Rightarrow\left|x+3\right|=28\)
\(\Rightarrow\left[{}\begin{matrix}x+3=28\\x+3=-28\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=28-3=25\\x=-28-3=-31\end{matrix}\right.\)
c) \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Rightarrow2-\left(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}\right)=2-\left(\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\right)\)
\(\Rightarrow2-\dfrac{x-1}{2009}-\dfrac{x-2}{2008}=2-\dfrac{x-3}{2007}-\dfrac{x-4}{2006}\)
\(\Rightarrow1-\dfrac{x-1}{2009}+1-\dfrac{x-2}{2008}=1-\dfrac{x-3}{2007}+1-\dfrac{x-4}{2006}\)
\(\Rightarrow\dfrac{2009-\left(x-1\right)}{2009}+\dfrac{2008-\left(x-2\right)}{2008}=\dfrac{2007-\left(x-3\right)}{2007}+\dfrac{2006-\left(x-4\right)}{2006}\)
\(\Rightarrow\dfrac{2009-x+1}{2009}+\dfrac{2008-x+2}{2008}=\dfrac{2007-x+3}{2007}+\dfrac{2006-x+4}{2006}\)
\(\Rightarrow\dfrac{2010-x}{2009}+\dfrac{2010-x}{2008}=\dfrac{2010-x}{2007}+\dfrac{2010-x}{2006}\)
\(\Rightarrow\dfrac{2010-x}{2009}+\dfrac{2010-x}{2008}-\dfrac{2010-x}{2007}-\dfrac{2010-x}{2006}=0\)
\(\Rightarrow\left(2010-x\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Mà \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)
=> 2010 - x = 0
=> x = 2010 - 0
=> x = 2010