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b) Ta có: \(5-\left|3x-1\right|=3\)
\(\Leftrightarrow\left|3x-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-2\\3x-1=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=1\end{matrix}\right.\)
Vậy: \(x\in\left\{-\frac{1}{3};1\right\}\)
c) Ta có: \(\left(1-2x\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}1-2x=3\\1-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;2\right\}\)
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201410 + 20149 = 20149[2014 + 1] = 20149.2015
201510 = 2015.20159
Vì 20149.2015 < 2015.20159
Nên 201410 + 20149 < 201510
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a) 92x : 9 =729 <---> 92x=729.9 <--->92x=6561 <---->92x=92.2--->x=2
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đặt tổng trên là A
ta co:
1/10^2<1/9.10
1/11^2<1/10.11
........
1/2014^2
=>A<1/9.10+1/10.11+.......+1/2013.2014=1/9-1/2014<1/9<9(đpcm)
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cùng nhân tử với 2014>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
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⇔2014-3x-9=20140
⇒-3x-9=0
⇔-3x=9
⇔x=-3
Vậy x=-3
2014-3x-9 = 1
2014-3x9 = 20140
-3x-9 = 0
-3x = 9
x = -3