x = 184

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\(\pm3\)

\(\frac{|x|}{186}=\left(1-\frac{30}{31}\right)+\left(\frac{60}{61}-1\right)\)

\(\Leftrightarrow|x|=186\left(\frac{1}{31}-\frac{1}{61}\right)\)

\(\Leftrightarrow|x|=6-\frac{186}{61}\)

\(\Leftrightarrow|x|=\frac{180}{61}\)

\(\Leftrightarrow x=\pm\frac{180}{61}\)

\(\frac{|x|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)\)

\(\frac{|x|}{186}=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)\)

\(\frac{|x|}{186}=1-\frac{30}{31}+\frac{61}{62}-1\)

\(\frac{|x|}{186}=\left(1-1\right)+\left(\frac{61}{62}-\frac{30}{31}\right)\)

\(\frac{|x|}{186}=\frac{1}{62}\)

\(\Rightarrow|x|=186.\frac{1}{62}\)

\(|x|=3\)

\(\Rightarrow x=\pm3\)

\(1-\frac{303030}{313131}+\frac{616161}{626262}-1+\frac{929292}{939393}-1=\left(1-\frac{30}{31}\right)+\left(\frac{61}{62}-1\right)+\left(\frac{92}{93}-1\right)\)

\(=\frac{1}{31}-\frac{1}{62}-\frac{1}{93}=\frac{1}{186}\)

=> \(\frac{\left|x\right|}{186}=\frac{1}{186}\)=> |x| = 1 => x = 1 hoặc x = - 1

Vậy...............

303030/313131=30/31(1)

616161/626262=61/62(2)

929292/939393=92/92(3)

từ 1,2,3=>tự làm

phải làm mới đc **** chứ !

\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}+\frac{x+1}{14}=0\)

\(\left(x+1\right).\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\)\(\ne\)0 nên x + 1 = 0 \(\Rightarrow\)x = -1

\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x.\left(x+1\right)}=\frac{2011}{2013}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2013}:2\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)

\(\frac{\left(x+1-2\right)}{2.\left(x+1\right)}=\frac{2011}{4026}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

=> \(2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)

=> \(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2019}{2020}\)

=> \(1-\frac{2}{x+1}=\frac{2019}{2020}\)

=> \(\frac{2}{x+1}=\frac{1}{2020}=\frac{2}{4040}\)

=> x + 1 = 4040 => x = 4039

-1

​giúp con với ông bà các bác ơi

a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)

\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)

b.

\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)

\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)

\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)

\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)

\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)