\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)

\(S=\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}\right)-\left(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+\dfrac{1}{8.10}\right)\)

\(S=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(S=\dfrac{1}{2}-\dfrac{1}{18}-\dfrac{1}{4}+\dfrac{1}{20}\)

\(S=.C.A.S.I.O.\)

\(S=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+\dfrac{1}{7\cdot9}-\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+\dfrac{1}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}\right)-\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+\dfrac{2}{8\cdot10}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{9}\right)-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{8}{9}-\dfrac{1}{2}\cdot\dfrac{2}{5}\)

\(=\dfrac{4}{9}-\dfrac{1}{5}\)

\(=\dfrac{11}{45}\)

`@` `\text {Ans}`

`\downarrow`

\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=\dfrac{2^3}{1}=2^3=8\)

_____

\(\dfrac{4^8\cdot9^4}{6^6\cdot8^3}\)

`=`\(\dfrac{\left(2^2\right)^8\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)

`=`\(\dfrac{2^{16}\cdot3^8}{2^{15}\cdot3^6}\)

`=`\(\dfrac{3^2}{2}\) `=`\(\dfrac{9}{2}\)

______

\(\dfrac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}\)

`=`\(\dfrac{\left(3^3\right)^4\cdot2^3-3^{10}\cdot\left(2^2\right)^3}{2^4\cdot3^4\cdot\left(3^2\right)^3}\)

`=`\(\dfrac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^4\cdot3^4\cdot3^6}\)

`=`\(\dfrac{3^{10}\cdot\left(3^2\cdot2^3-2^6\right)}{3^{10}\cdot2^4}\)

`=`\(\dfrac{72-2^6}{2^4}=\dfrac{8}{16}=\dfrac{1}{2}\)

\(\dfrac{2^8-2^3}{2^5-1}=\dfrac{2^3\left(2^5-1\right)}{2^5-1}=2^3=8\)

\(\dfrac{4^8.9^4}{6^6.8^3}=\dfrac{2^{16}.3^8}{2^6.3^6.2^9}=2.3^2=18\)

\(\dfrac{27^4.2^3-3^{10}.4^3}{6^4.9^3}=\dfrac{3^{12}.2^3-3^{10}.2^6}{2^4.3^4.3^6}=\dfrac{2^3.3^{10}.\left(3^2-2^3\right)}{2^4.3^{10}}=\dfrac{9-8}{2}=\dfrac{1}{2}\)

\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{48.50}\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{48.50}\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{50-48}{48.50}\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)\)

\(\Leftrightarrow\dfrac{1}{2}.\dfrac{24}{50}=\dfrac{6}{25}\)

=1/2 - 1/4 + 1/4 - 1/6 + ... + 1/98 - 1/100

=1/2 - 1/100 = 49/100

1/2 - 1/4 +  1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100

= 1/2 - 1/100 

= 49/100

1/2.4 + 1/4.6 + 1/6.8 + ... + 1/96.98 + 1/98.100

= 1/2 . ( 2/2.4 + 2/4.6 + 2/6.8 + ... + 2/96.98 + 2/98.100 )

= 1/2 . ( 1/2 - 1/4 + 1/4 - 1/6 + 1/6 - 1/8 + ... + 1/96 - 1/98 + 1/98 - 1/100 )

= 1/2 . ( 1/2 - 1/100 )

= 1/2 . ( 50/100 - 1/100 )

= 49/200

1)

a) \(|x-3,5|=7,5\)

\(\Rightarrow x-3,5=7,5\)

hay \(x-3,5=-7,5\)

TH1 : \(x-3,5=7,5\Rightarrow x=7,5+3,5=11\)

TH2 : \(x-3,5=-7,5\Rightarrow x=-7,5+3,5=-4\)

b) \(|x+\dfrac{4}{5}|-\dfrac{1}{2}=0\)

\(\Rightarrow\left(x+\dfrac{4}{5}\right)-\dfrac{1}{2}=0\) (chỉ có 1 TH vì số 0 ko phải dương or âm)

\(\left(x+\dfrac{4}{5}\right)=0+\dfrac{1}{2}=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{4}{5}=\dfrac{5-8}{10}=\dfrac{-3}{10}\)

c) \(3,6-|x-0,4|=0\)

\(\Rightarrow3,6-\left(x-0,4\right)=0\) ( giải thích giống câu b )

\(\Rightarrow-\left(x-0,4\right)=0-3,6\)

\(\Rightarrow-\left(x-0,4\right)=-3,6\)

\(\Rightarrow-x+0,4=-3,6\) ( Phá dấu )

\(\Rightarrow-x=-3,6-0,4=-3,6+\left(-0,4\right)=-4\)

\(\Rightarrow x=4\)

d) \(-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)

\(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{-5}{9}\)

hay \(\Rightarrow-\dfrac{5}{12}:|\dfrac{-5}{6}:x|=\dfrac{5}{9}\)

TH1 : \(-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{-5}{9}\Rightarrow\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\left(-\dfrac{5}{9}\right)\)

\(\Rightarrow\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{12}.\dfrac{9}{5}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{5}{6}:\dfrac{3}{4}=-\dfrac{5.4}{6.3}=-\dfrac{5.2}{3.3}=-\dfrac{10}{9}\)

TH2 : \(\Rightarrow-\dfrac{5}{12}:\left(-\dfrac{5}{6}:x\right)=\dfrac{5}{9}\)

\(\Rightarrow\)\(\left(-\dfrac{5}{6}:x\right)=-\dfrac{5}{12}:\dfrac{5}{9}=-\dfrac{5.9}{12.5}=-\dfrac{9}{12}=-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{5}{6}:\left(-\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{4}{3}=\dfrac{10}{9}\)

Vậy x = ....

e)

Vì \(|x-3,5|\ge0;|4,5-x|\ge0\) với mọi x

Do đó : \(|x-3,5|+|4,5-x|=0\)

\(\Rightarrow|x-3,5|=0;|4,5-x|=0\)

\(\Rightarrow x-3,5=0\) và \(4,5-x=0\)

\(\Rightarrow x=0+3,5=3,5\) và \(-x=0+4,5=4,5\Rightarrow x=-4,5\)

( không đồng thời xảy ra)

\(\Rightarrow\) Không tồn tại x thuộc Q để \(|x-3,5|+|4,5-x|=0\)

2)

a) Đề sai

b) (45,3 + 7,3) + (-22)

= 52,6 + (-22) = 30,6

c) [(-11.7) + (11.7)] + [5.5+10]

= 0 + 15.5 = 15.5

( Câu c bạn cho rối quá )

d) [(-6.8) + 2.8] + [(-56.9) + 5.9 ]

= (-4) + (-51) = 55

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)

p: \(F=\dfrac{1}{3}\left(\dfrac{3}{3\cdot6}+\dfrac{3}{6\cdot9}+\dfrac{3}{9\cdot12}+...+\dfrac{3}{30\cdot33}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{30}-\dfrac{1}{33}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{10}{33}=\dfrac{10}{99}\)

n: \(F=2\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2008}-\dfrac{1}{2010}\right)\)

\(=2\cdot\dfrac{502}{1005}=\dfrac{1004}{1005}\)

m: \(=\left(3-\dfrac{7}{3}+\dfrac{1}{4}\right):\left(4-\dfrac{31}{6}+\dfrac{9}{4}\right)\)

\(=\dfrac{36-28+3}{12}:\dfrac{48-62+27}{12}\)

\(=\dfrac{11}{13}\)