`a, x-2=7/15`

`=>x=7/15 +2`

`=>x= 7/15+ 30/15`

`=>x= 37/15`

`b, 9/20 +x=2/5 +3/20`

`=> 9/20 +x=8/20 +3/20`

`=> 9/20 +x=11/20`

`=>x=11/20-9/20`

`=>x= 2/20=1/10`

x = \(\dfrac{7}{15}\) + 2 = \(\dfrac{37}{15}\) 

x = \(\dfrac{2}{5}\) + \(\dfrac{3}{20}\) - \(\dfrac{9}{20}\) = \(\dfrac{1}{10}\)

bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)

   \(x=\dfrac{7}{35}+\dfrac{-15}{35}\)

   \(x=-\dfrac{8}{35}\)

\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}:\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}\times\dfrac{2}{3}\)

                 \(x=\dfrac{8}{21}\)

\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)

   \(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

           \(x=-\dfrac{7}{6}-\dfrac{3}{4}\)

           \(x=-\dfrac{23}{12}\)

\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)

    \(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)

     \(\dfrac{-5}{9}-x=\dfrac{13}{18}\)

                \(x=\dfrac{-5}{9}-\dfrac{13}{18}\)

                \(x=\dfrac{-10}{18}-\dfrac{13}{18}\)

                \(x=-\dfrac{23}{18}\)

a, 2/5 + 3/4 : x = -1/2

3/4 : x = -1/2 - 2/5

3/4 : x = -9/10

x = 3/4 : -9/10

x = -5/6

b, 5/7 - 2/3 . x = 4/5 

2/3 . x = 4/5 + 5/7

2/3 . x = 53/35

x = 53/35 : 2/3

x = 159/70

a) (-5/9)^10 : x = (-5/9)^8

=> x = (-5/9)^10 : (-5/9)^8

=> x = (-5/9)^10-8 = (-5/9)^2

=> x = 25/81

b ) x : (-5/9)^8 = (-9/5)^8

=> x = (-9/5)^8 . (-5/9)^8

=> x = ( (-9)^8.(-5)^8 )/(5^8 . 9^8 )

=> x = 1

C) x^3 = -8 =(-2)^3

=> x = -2

a) (-5/9)¹⁰ : x = (-5/9)⁸

x = (-5/9)¹⁰ : (-5/9)⁸

x = (-5/9)²

x = 25/81

b) x : (-5/9)⁸ = (-9/5)⁸

x = (-9/5)⁸ . (-5/9)⁸

x = [-9/5 . (-5/9)]⁸

x = 1⁸

x = 1

c) x³ = -8

x³ = (-2)³

x = -2

chắc đéo biết

pro tìm ra x rồi còn j

a: Ta có: \(\dfrac{1}{4}:x=3\dfrac{4}{5}:40\dfrac{8}{15}\)

\(\Leftrightarrow x=\dfrac{1}{4}\cdot\dfrac{\dfrac{608}{15}}{3+\dfrac{4}{5}}\)

\(\Leftrightarrow x=\dfrac{152}{15}:\dfrac{19}{5}=\dfrac{8}{3}\)

b: Ta có: \(\left(x+1\right):\dfrac{5}{6}=\dfrac{20}{3}\)

\(\Leftrightarrow x+1=\dfrac{50}{9}\)

hay \(x=\dfrac{41}{9}\)

c: Ta có: \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

\(\Leftrightarrow x^2-1=63\)

\(\Leftrightarrow x^2=64\)

hay \(x\in\left\{8;-8\right\}\)

c. \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\) 

    \(7.9=\left(x-1\right).\left(x+1\right)\) 

    \(63=x^2-1\) 

             \(x^2=63+1\) 

             \(x^2=64\) 

             \(x^2=8^2\)

             \(x=8\)          

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)

\(a, x^3+5x^2-9x-45=0\\ \Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\left(x\ne-5\right)\\ \text{Với }x=3\Leftrightarrow A=\dfrac{9-9}{3\left(3+5\right)}=0\\ \text{Với }x=-3\Leftrightarrow A=\dfrac{9-9}{3\left(-3+5\right)}=0\\ \text{Vậy }A=0\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)