`250:(x-8)=25`

`=> x-8=250:25`

`=> x-8=10`

`=> x=10+8`

`=> x=18`

250 : ( x - 8 ) = 25

x - 8 = 250 : 25

x - 8 = 10

x = 10 + 8

x = 18

vậy x = 18

( \(\dfrac{3}{25}\) - \(\dfrac{4}{5}\)) x \(\dfrac{25}{3}\)

 = ( \(\dfrac{3}{25}\) - \(\dfrac{20}{25}\)) x \(\dfrac{25}{3}\)

= \(-\dfrac{17}{25}\) x \(\dfrac{25}{3}\)

= - \(\dfrac{17}{3}\)

`(3/25 - 4/5) . 25/3`

`=(3/25 - 20/25) . 25/3`

`= -17/25 . 25/3`

`= -17/3`

ĐK : \(x;y\ne0\)

Ta có : \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x+y}{xy}=\dfrac{1}{8}\)

\(\Leftrightarrow8.\left(x+y\right)=xy\)

\(\Leftrightarrow xy-8x-8y=0\)

\(\Leftrightarrow\left(xy-8x\right)-\left(8y-64\right)=64\)

\(\Leftrightarrow x.\left(y-8\right)-8.\left(y-8\right)=64\Leftrightarrow\left(x-8\right).\left(y-8\right)=64\)

Do x;y \(\inℤ\) ta có bảng sau

Vậy (x;y) = (9;72) ; (10  ; 40) ; (12 ; 24) ; (16;16) ; (24;12) ; (7;-56) ; (6;-24) ; (4;-8) và các hoán vị của chúng 

a) đặt

 \(S=1+\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\\ 2S=2+\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{99\cdot101}\\ 2S=2+\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ 2S=2+1-\dfrac{1}{101}\\ 2S=\dfrac{302}{101}\\ S=\dfrac{151}{101}\)

b)

đặt

\(S=\dfrac{1}{2}+\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{98\cdot101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{98}-\dfrac{1}{101}\\ 3S=\dfrac{3}{2}+\dfrac{1}{2}-\dfrac{1}{101}\\ 3S=\dfrac{201}{101}\\ S=\dfrac{67}{101}\)

\(2A-1=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(2A-1=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(2A=\dfrac{201}{101}\Rightarrow A=\dfrac{201}{202}\)

\(\dfrac{n}{n-5}=\dfrac{n-5+5}{n-5}=\dfrac{n-5}{n-5}+\dfrac{5}{n-5}=1+\dfrac{5}{n-5}\)

Để \(\dfrac{n}{n-5}\in Z\) thì \(\dfrac{5}{n-5}\in Z\)

\(\Rightarrow n-5\in\text{Ư}_{\left(5\right)}=\left\{-5;-1;1;5\right\}\)

\(\Rightarrow n\in\left\{0;4;6;10\right\}\)

Do \(n\in N\) nên tất cả các giá trị đều nhận

Vậy ...

`(x-3/8) +1/4= 7/8`

`=> x-3/8 = 7/8-1/4`

`=> x-3/8 = 7/8-2/8`

`=> x-3/8 =5/8`

`=>x= 5/8 +3/8`

`=>x= 8/8=1`

khó thế =))

\(S=\dfrac{2^2}{3x5}+\dfrac{2^2}{5x7}+\dfrac{2^2}{7x9}+...+\dfrac{2^2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{2}{3x5}+\dfrac{2}{5x7}+\dfrac{2}{7x9}+...+\dfrac{2}{97x99}\)

\(\dfrac{S}{2}=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)

S=\(\dfrac{64}{99}\)