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\(\frac{\left|x\right|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929291}{939393}-1\right)\)
<=> \(\frac{\left|x\right|}{186}=-\frac{303030}{313131}+\frac{616161}{626262}+\frac{929292}{939393}+1-1-1\)
<=> \(\frac{\left|x\right|}{186}=-\frac{30}{31}+\frac{61}{62}+\frac{92}{93}+1-1-1\)
<=> \(\frac{\left|x\right|}{186}=\frac{61}{62}+\frac{92}{93}-1-\frac{30}{31}\)
<=> \(\frac{\left|x\right|}{186}=-\frac{1.31}{31}+\frac{61}{62}+\frac{92}{91}-\frac{30}{31}\)
Lấy MSC là 168, ta có:
<=> \(\frac{\left|x\right|}{186}=\frac{-186}{186}+\frac{183}{186}+\frac{184}{186}-\frac{180}{186}\)
<=> \(\frac{\left|x\right|}{186}=\frac{-186+183+184-180}{186}\)
<=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)
<=> |x| = 1
<=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
\(1-\frac{303030}{313131}+\frac{616161}{626262}-1+\frac{929292}{939393}-1=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)+\left(\frac{92}{93}-1\right)\)
\(=\frac{1}{31}-\frac{1}{62}-\frac{1}{93}=\frac{1}{186}\)
=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)=> |x| = 1 => x = 1 hoặc x = - 1
Vậy...............
Bạn xét 3 TH nha :
x _< 1
1< x< 3
x >_ 3 là ra
tick luôn nha
1-2x =8
=> 2x = 1-8
=> 2x = -7
=>x =-7 : 2
=> x = -3,5
Vậy tac có:
2x-1+ /-3,5/
= 2x-1+ 3,5 = 2x- 4,5
ko hiểu đè bài lám nên làm sai bỏ qua nhá :>>>
Ta có: |2x-1|+|1-2x|=8
\(\Leftrightarrow\left|2x-1\right|+\left|2x-1\right|=8\)(Vì 2x-1 và 1-2x là hai số đối nhau)
\(\Leftrightarrow2\left|2x-1\right|=8\)
\(\Leftrightarrow\left|2x-1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=4\\2x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{2};-\dfrac{3}{2}\right\}\)
X=2008;2009;2010;2011;2012;2013;2014;2015;2016;2017;.............................................
ngoăcj vuông là j vậy
\(\dfrac{x}{186}=\left(1-\dfrac{3030}{3131}\right)+\left(\dfrac{6161}{6262}-1\right)+\left(\dfrac{929292}{939393}-1\right)\\ \dfrac{x}{186}=\left(1-\dfrac{30}{31}\right)+\left(\dfrac{61}{62}-1\right)+\left(\dfrac{92}{93}-1\right)\\ \dfrac{x}{186}=\dfrac{1}{31}+\dfrac{-1}{62}+\dfrac{-1}{93}\\ \dfrac{x}{186}=\dfrac{6}{186}+\dfrac{-3}{186}+\dfrac{-2}{186}\\ \dfrac{x}{186}=\dfrac{1}{186}\\ \Rightarrow x=1\\ \Rightarrow\left|x\right|=1\)
Vậy \(\left|x\right|=1\)