\(\frac{18}{2.5}+\frac{18}{5.8}+....+\frac{18}{103.106}\)

=\(6\left(\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{103.106}\right)\)

=\(6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+....+\frac{1}{103}-\frac{1}{106}\right)\)

=\(6\left(\frac{1}{2}-\frac{1}{106}\right)\)

=\(6.\frac{26}{53}\)

=\(\frac{156}{53}\)

=6.3/2.5 +6.3/5.8+...+6.3/203.206

=6(3/2.5+3/5.8+...+3/203.206)

=6(1/2-1/5+1/5-1/8+...+1/203-1/206)

=6[(1/2-1/206)+(1/5-1/5)+(1/8-1/8)+...+(1/203-1/203)]

=6(1/2-1/206)=6(103/206-1/206)=6. 102/206=6. 51/103=306/103

A=6.( 3/2.5+3/5.8+...+3/203.206)

  =6.(1/2-1/5+1/3-1/8+...+1/202-1/206)

  =6.(1/2-1/206)=306/103

A = \(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

   = \(18.\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{203.206}\right)\)

   = \(\frac{18}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

   = \(6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\) 

  = \(6.\left(\frac{1}{2}-\frac{1}{206}\right)\)

  = \(6.\frac{51}{103}\)

  = \(2\frac{100}{103}\)

\(A=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)

\(A=\dfrac{6.3}{2.5}+\dfrac{6.3}{5.8}+...+\dfrac{6.3}{203.206}\)

\(A=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)

\(A=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)

\(A=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)

\(A=6.\dfrac{51}{103}\)

\(A=\dfrac{306}{103}\)

18/(2.5) + 18/(5.8) + .... + 18/(203.206)

= 18.[1/(2.5) + 1/(5.8) + .... + 1/(203.206)]

= 18.(1/2 - 1/5 + 1/5 - 1/8 + .... + 1/203 - 1/206)

=18.(1/2 - 1/206)

=18.(51/103)

=918/103

a) \(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=4.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=4.\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=4.\frac{32}{99}\)

\(=\frac{128}{99}\)

\(\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{97.99}\)

\(=2\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=2\left(\frac{1}{3}-\frac{1}{99}\right)\)

\(=2.\frac{32}{99}\)

\(=\frac{64}{99}\)

\(\frac{18}{2.5}+\frac{18}{5.8}+...+\frac{18}{203.206}\)

\(=6\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{203.206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{203}-\frac{1}{206}\right)\)

\(=6\left(\frac{1}{2}-\frac{1}{206}\right)\)

\(=6.\frac{102}{206}\)

\(=\frac{612}{206}\) ( tự rút gọn

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{18}\)

`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{18}\right)\)

`=`\(\dfrac{7}{5}\cdot\left(\dfrac{20}{19}-\dfrac{1}{18}\right)\)
`=`\(\dfrac{7}{5}\cdot\dfrac{341}{342}=\dfrac{2387}{1710}\)

{[83 - (4².10+2².5):18] + 27}.3² + 100 = {[83 - 180:18]+27}.9 + 100 = {73 + 27}.9 +100 = 900 + 100 = 1000

{[83 - (4^2.10 + 2^2.5):18]+27}.3^2 + 100 = [83- 10 + 27].3^2 = 100.9 = 900