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21 tháng 9 2015
\(1-\frac{303030}{313131}+\frac{616161}{626262}-1+\frac{929292}{939393}-1=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)+\left(\frac{92}{93}-1\right)\)
\(=\frac{1}{31}-\frac{1}{62}-\frac{1}{93}=\frac{1}{186}\)
=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)=> |x| = 1 => x = 1 hoặc x = - 1
Vậy...............
12 tháng 8 2019
\(\frac{|x|}{186}=\left(1-\frac{30}{31}\right)+\left(\frac{60}{61}-1\right)\)
\(\Leftrightarrow|x|=186\left(\frac{1}{31}-\frac{1}{61}\right)\)
\(\Leftrightarrow|x|=6-\frac{186}{61}\)
\(\Leftrightarrow|x|=\frac{180}{61}\)
\(\Leftrightarrow x=\pm\frac{180}{61}\)
NT
2
MV
22 tháng 7 2017
\(\dfrac{x}{186}=\left(1-\dfrac{3030}{3131}\right)+\left(\dfrac{6161}{6262}-1\right)+\left(\dfrac{929292}{939393}-1\right)\\ \dfrac{x}{186}=\left(1-\dfrac{30}{31}\right)+\left(\dfrac{61}{62}-1\right)+\left(\dfrac{92}{93}-1\right)\\ \dfrac{x}{186}=\dfrac{1}{31}+\dfrac{-1}{62}+\dfrac{-1}{93}\\ \dfrac{x}{186}=\dfrac{6}{186}+\dfrac{-3}{186}+\dfrac{-2}{186}\\ \dfrac{x}{186}=\dfrac{1}{186}\\ \Rightarrow x=1\\ \Rightarrow\left|x\right|=1\)
Vậy \(\left|x\right|=1\)
10 tháng 9 2021
\(B=\dfrac{1+\dfrac{1}{7}+\dfrac{1}{7^2}-\dfrac{1}{7^3}}{4+\dfrac{4}{7}+\dfrac{4}{7^2}-\dfrac{4}{7^3}}\cdot\dfrac{858585}{313131}\cdot\left(-1\dfrac{14}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{85}{31}\cdot\dfrac{-31}{17}\)
\(=\dfrac{-5}{4}\)
10 tháng 11 2019
a) 74x.(3312+33332020+333333303030+3333333342424242)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)=3247x.(1233+20203333+303030333333+4242424233333333)=32
74x.(3312+3320+3330+3342)=32\frac{7}{4}x.\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)=3247x.(1233+2033+3033+4233)=32
74x.(333.4+334.5+335.6+336.7)=32\frac{7}{4}x.\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)=3247x.(3.433+4.533+5.633+6.733)=32
74x.33.(13−14+14−15+15−16+16−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)=3247x.33.(31−41+41−51+51−61+61−71)=32
74x.33.(13−17)=32\frac{7}{4}x.33.\left(\frac{1}{3}-\frac{1}{7}\right)=3247x.33.(31−71)=32
74x.33⋅421=32\frac{7}{4}x.33\cdot\frac{4}{21}=3247x.33⋅214=32
10 tháng 11 2019
b) 13+16+110+115+...+2x.(x−1)=20072009\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{2}{x.\left(x-1\right)}=\frac{2007}{2009}31+61+101+151+...+x.(x−1)2=20092007
26+212+220+230+...+2(x−1).x=20072009\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}62+122+202+302+...+(x−1).x2=20092007
22.3+23.4+24.5+25.6+...+2(x−1).x=20072009\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{\left(x-1\right).x}=\frac{2007}{2009}2.32+3.42+4.52+5.62+...+(x−1).x2=20092007
2.(12−13+13−14+14−15+15−16+...+1x−1−1x)=200720092.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x-1}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−31+31−41+41−51+51−61+...+x−11−x1)=20092007
2.(12−1x)=200720092.\left(\frac{1}{2}-\frac{1}{x}\right)=\frac{2007}{2009}2.(21−x1)=20092007
1−2x=200720091-\frac{2}{x}=\frac{2007}{2009}1−x2=20092007
2x=22009\frac{2}{x}=\frac{2}{2009}x2=20092
=> x = 2009
\(\frac{\left|x\right|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929291}{939393}-1\right)\)
<=> \(\frac{\left|x\right|}{186}=-\frac{303030}{313131}+\frac{616161}{626262}+\frac{929292}{939393}+1-1-1\)
<=> \(\frac{\left|x\right|}{186}=-\frac{30}{31}+\frac{61}{62}+\frac{92}{93}+1-1-1\)
<=> \(\frac{\left|x\right|}{186}=\frac{61}{62}+\frac{92}{93}-1-\frac{30}{31}\)
<=> \(\frac{\left|x\right|}{186}=-\frac{1.31}{31}+\frac{61}{62}+\frac{92}{91}-\frac{30}{31}\)
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
<=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)