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giống cái kia thôi bn
Mik làm rồi mà
Mà cái bn Nguyễn Duy Đạt gì đó làm thiếu 1 trường hợp
Mà bn vẫn kik hở
Sao zzzzz??????
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(2x-1)2 = (2x-1)6
(2x-1)2 - (2x-1)6 = 0
(2x-1)2 x (1-(2x-1)4) = 0
=> 2x-1 = 0 <=> x=1/2
=> 2x-1=1 <=> x=1
vội cũng phải cho mik nha
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a/ \(\left|1-2x\right|>7\Leftrightarrow\left[{}\begin{matrix}1-2x=7\\1-2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x< -6\\2x< 8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< -3\\x< 4\end{matrix}\right.\)
b/ \(\dfrac{-5}{x-3}< 0\Leftrightarrow x-3>0\) ( vì -5<0)
\(\Leftrightarrow x>3\)
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\(\left(2x+1\right)^3=-0,001\)
\(\Rightarrow\left(2x+1\right)^3=\left(-0,1\right)^3\)
\(\Rightarrow2x+1=-0,1\)
\(\Rightarrow2x=-\frac{1}{10}-1\)
\(\Rightarrow2x=-\frac{11}{10}\)
\(\Rightarrow x=-\frac{11}{10}:2\)
\(\Rightarrow x=-\frac{11}{10}.\frac{1}{2}=-\frac{11}{20}\)
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\(16\left(2x-1\right)^{2010}=\left(1-2x\right)^{2014}\)
\(\Rightarrow\)\(2^4\left(2x-1\right)^{2010}=\left(1-2x\right)^{2014}\) \(\Leftrightarrow\) \(2^4\left(2x-1\right)^{2010}=\left(1-2x\right)^{2010}.\left(1-2x\right)^4\)
\(\Rightarrow\)\(\hept{\begin{cases}2x-1=2\\1-2x=\left(-2\right)\end{cases}}\) ( Hai vế trái dấu )
\(\Rightarrow\) \(\hept{\begin{cases}x=\frac{3}{2}\\x=\frac{3}{2}\end{cases}}\)
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a) [2x] = -1\(\Rightarrow-1\le2x< 0\Rightarrow-0,5\le x< 0\)
b) [x + 0,4] = 3\(\Rightarrow3\le x+0,4< 4\Rightarrow2,6\le x< 3,6\)
c)\(\left[\frac{2}{3}x-5\right]=3\Rightarrow3\le\frac{2}{3}x-5< 4\Rightarrow8\le\frac{2}{3}x< 9\Rightarrow12\le x< 13,5\)
Từ bài trên,ta có :\(\left[x\right]=y\Rightarrow y\le x< y+1\left(x\in Q;y\in Z\right)\)
\(\left(2x+1\right)^5=\left(2x+1\right)^{2010}\)
\(\Rightarrow\left(2x+1\right)^{2010}-\left(2x+1\right)^5=0\)
\(\Rightarrow\left(2x+1\right)^5.\left[\left(2x+1\right)^{2005}-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x+1\right)^5=0\\\left(2x+1\right)^{2005}-1=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x+1=0\\\left(2x+1\right)^{2005}=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x+1=0\\2x+1=1\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=-1\\2x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=-\frac{1}{2}\\x=0\end{cases}}\)
(2x+1)5=(2x+1)2010
=> 2x+1=1 hoặc 2x+1=0
=>2x=0 hoặc 2x=1
=>x=0 hoặcx=0,5