x/186 = (1 - 303030/313131) + (616161/626262 - 1) + (929292/939393 - 1)

x/186 = 1 - 30/31 + 61/62 - 1 + 92/93 - 1

x/186 = 61/62 - 30/31 + 92/93 - 1

x/186 = 1/62 - 1/93

x/186 = 1/186

x = 1

x = 184

xim may ban trnh bay chi tiet duoc khong 

cam on rat nhieu

\(\pm3\)

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

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

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

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

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

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

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

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

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

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

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

\(|x|=3\)

\(\Rightarrow x=\pm3\)

\(\frac{\left|x\right|}{186}=\left(1-\frac{303030}{313131}\right)+\left(\frac{616161}{626262}-1\right)+\left(\frac{929291}{939393}-1\right)\)

<=> \(\frac{\left|x\right|}{186}=-\frac{303030}{313131}+\frac{616161}{626262}+\frac{929292}{939393}+1-1-1\)

<=> \(\frac{\left|x\right|}{186}=-\frac{30}{31}+\frac{61}{62}+\frac{92}{93}+1-1-1\)

<=> \(\frac{\left|x\right|}{186}=\frac{61}{62}+\frac{92}{93}-1-\frac{30}{31}\)

<=> \(\frac{\left|x\right|}{186}=-\frac{1.31}{31}+\frac{61}{62}+\frac{92}{91}-\frac{30}{31}\)

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

<=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

x = từ 1 đến 10000....0

Nhận xét :

\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

Vì \(x\ge0\) nên pt a) tương đương với : \(100x+\frac{1+2+3+...+100}{101}=101x\)

\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)

b) 

Tương tự câu a) , phương trình tương đương với : 

\(49x+\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{...1}{97.99}=50x\)

\(\Rightarrow x=\frac{97}{195}\)

c) 

Tương tự câu a) , phương trình tương đương với : 

\(99x+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}=100x\)

\(\Rightarrow x=\frac{99}{100}\)

\(\frac{1}{\left(x-1\right)x}+\frac{1}{\left(x-2\right)\left(x-1\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}=\frac{x}{x^2-4x}\)

\(\Leftrightarrow\)\(\frac{1}{x-1}-\frac{1}{x}+\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{x}{x\left(x-4\right)}\)

\(\Leftrightarrow\)\(-\frac{1}{x}+\frac{1}{x-4}=\frac{1}{x-4}\)

\(\Leftrightarrow\)\(\frac{-\left(x-4\right)+x}{x\left(x-4\right)}=\frac{x}{x\left(x-4\right)}\)

\(\Leftrightarrow\)\(4-x+x=x\)

\(\Leftrightarrow x=4\)

lo nói mk làm cách lâu chứ m cx hỏi người khác!!!!!!!!!!!