\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\)

\(S=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\right)\)

\(S=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{9900}\right)\)

\(S=\frac{1}{2}\cdot\frac{4949}{9900}=\frac{4949}{19800}\)

\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\)

\(\Rightarrow2S=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{98.99.100}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)

\(=\frac{1}{1.2}-\frac{1}{99.100}=\frac{4849}{9900}\)

\(\Rightarrow S=\frac{4949}{9900}\div2=\frac{4949}{19800}\)

Ta có : \(2\le\left|x\right|\le5\)

\(\Rightarrow x\in\left\{\pm3;\pm4\right\}\)

Vậy \(x\in\left\{\pm3;\pm4\right\}\)

Cho tớ xin lỗi, tớ thiếu số \(\pm2\)và \(\pm5\)

45-(x-9)=-35  

=> x - 9 = 45 - ( -35 ) 

=> x - 9 = 80

=> x = 80 + 9 

=> x = 89 

 Vậy x = 89

45-(x-9)=-35

      x-9=45+35

      x-9=80

      x=80+9

      x=89

Ta có : 

 sao chỗ này bn ghi x= 3 thế  là - hay + để mk còn giải

=312.(-18)-312.81-312.1

=312.[(-18)-81-1]

=312.(-100)

=-31200

312.(-18)+(-312).81-312

=312.(-18)+312.(-81)-312

=312.[(-18)+(-81)-1]

=312.[(-99)-1]

=312.(-100)

=-31200

\(\left(2x+6\right)\left(5-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+6=0\\5-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=-6\\x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=5\end{cases}}\)

vậy x = -3 hoặc x=5

chắc hiểu liền

  ta có  (-x+3)=0 hoặc (x-2)=0

            (-x+3)=0

           - x=0-3=-3 suy ra x=3

           x-2=0 

           x=0+2=2

   đúng 100% luôn