\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)

\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times99\times5}+\frac{1}{99\times100\times5}\)

\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{1}{5}\times\frac{99}{100}=\frac{99}{500}\)

\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)

\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times90\times5}+\frac{1}{90\times100\times5}\)

\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{99}{500}\)

a) \(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}=4\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{98}-1+\frac{x-3}{97}-1+\frac{x-3}{96}-1=4-4\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\right)=0\)

\(\Rightarrow x-1=0\) ( vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne0\) )

Vậy x = 1

b) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=3\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=3-3\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}\ne0\)

=> x + 100 = 0

=> x           = -100

c) \(\frac{x-1}{99}+\frac{x-2}{49}+\frac{x-4}{32}=6\)

\(\Rightarrow\frac{x-1}{99}-1+\frac{x-2}{49}-2+\frac{x-4}{32}-3=6-6\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{49}+\frac{x-100}{32}=0\)

\(\Rightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\right)=0\)

Vì \(\frac{1}{99}+\frac{1}{49}+\frac{1}{32}\ne0\)

=> x - 100 = 0

=> x           = 100

Chúc bạn học tốt

có người khác trả lời trước rồi nên chị ko trả lời đâu nhé em trai

không biết giải

2001 

____

1991

39.(5 - 2\(x\)) = 39

     5 - 2\(x\)  = 39 : 39

    5 - 2\(x\) = 1

         2\(x\) = 5 - 1

         2\(x\) = 4

          \(x=4:2\)

           \(x=2\)

Vậy \(x=2\) 

\(x.1+x.2+x.3+....+x.99=495\)

<=> \(x.\left(1+2+3+...+99\right)=495\)

<=> \(x.\left\{\left(99+1\right).\left[\left(99-1\right):1+1\right]:2\right\}=495\)

<=> \(x.4950=495\)

<=> \(x=\frac{495}{4950}=\frac{1}{10}\)

x.1 +x.2+x.3+...+x.99=495             

=>x.(1+2+3+...+99)=495

=>x.\(\frac{\left(1+99\right).99}{2}=495\)

 =>x.4950=495

=>x=\(\frac{495}{4950}\)

=>x=1/10

a: \(\left(1-\frac12\right)\times\left(1-\frac13\right)\times\ldots\times\left(1-\frac{1}{99}\right)\)

\(=\frac12\times\frac23\times\ldots\times\frac{98}{99}\)

\(=\frac{1}{99}\)

b: \(\left(x+\frac12\right)+\left(x+\frac16\right)+\left(x+\frac{1}{12}\right)+\left(x+\frac{1}{20}\right)+\cdots+\left(x+\frac{1}{90}\right)=\frac{99}{10}\)

=>\(\left(x+\frac{1}{1\times2}\right)+\left(x+\frac{1}{2\times3}\right)+\cdots+\left(x+\frac{1}{9\times10}\right)=\frac{99}{10}\)

=>\(9\times x+1-\frac12+\frac12-\frac13+\cdots+\frac19-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x+1-\frac{1}{10}=\frac{99}{10}\)

=>\(9\times x=\frac{99}{10}+\frac{1}{10}-1=\frac{100}{10}-1=10-1=9\)

=>x=1