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a)\(\orbr{\begin{cases}5x-1=3\\5x-1=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}5x=4\\5x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=\frac{-2}{5}\end{cases}}\)
b) 2x+1=-0,1 <=> 2x=-1,1=>x=-0,55
c) (2x-3)4 .[1-(2x-3)2 ]=0
do (2x-3)4 lớn hơn 0 nên 1-(2x-3)2=0=>(2x-3)2=1=>2x-3=1=>2x=4=>x=2
d) tương tự câu c)
Tìm x
a) ( 2x + 1 )3 = - 0,001
b) ( 2x + 1 )5 = ( 2x + 1 )2016
Giúp mk vs, mk ko wên ơn dou! Help me...
a) (2x + 1)^3 = -0,001
(2x + 1)^3 = (-0,1)^3
=> 2x + 1 = -0,1
2x = -0,1 -1
2x = -1,1
x = (-1,1) : 2
x = -0,55
Vậy x = .......
( 2x + 1) 3 = -0,001
(2x+1)3=(-0,1)3
=>2x+1=-0,1
2x=-0,1-1
2x=\(\frac{-1}{10}-\frac{10}{10}\)
2x=\(\frac{-11}{10}\)
x=\(\frac{-11}{10}:2\)
x=\(\frac{-11}{10}.\frac{1}{2}\)
x=\(\frac{-11}{20}\)
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
\(\left(2-3x\right)^3.3^{12}=18^6\)
=>\(\left(2-3x\right)^3.3^{12}=2^6.\left(3^2\right)^6\)
=>\(\left(2-3x\right)^3.3^{12}=2^6.3^{12}\)
=>\(\left(2-3x\right)^3=2^6\)
=>\(\left(2-3x\right)^3=4^3\)
=>2-3x=4
=>3x=-2
=>x=\(\frac{-2}{3}\)
\(a)\) \(2^{333}\) và \(3^{222}\)
- Ta có: \(2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8< 9\) nên \(8^{111}< 9^{111}\) hay \(2^{333}< 3^{222}\)
Vậy \(2^{333}< 3^{222}\)
b: \(9^{1005}=3^{2010}>3^{2009}\)
c: \(99^{20}=99^{10}\cdot99^{10}\)
\(9999^{10}=99^{10}\cdot101^{10}\)
mà 99<101
nên \(99^{20}< 9999^{10}\)
d: \(\left(x-2013\right)^{2014}=1\)
=>x-2013=1 hoặc x-2013=-1
=>x=2014 hoặc x=2012
\(\left(2x+1\right)^3=-0,001\)
\(\Rightarrow\left(2x+1\right)^3=\left(-0,1\right)^3\)
\(\Rightarrow2x+1=-0,1\)
\(\Rightarrow2x=-0,1-1\)
\(\Rightarrow2x=-1,1\)
\(\Rightarrow x=\dfrac{-1,1}{2}\)
\(\Rightarrow x=-0,55\)