=>\(1\cdot\dfrac{2}{4}\cdot\dfrac{3}{6}\cdot...\cdot\dfrac{31}{62}\cdot\dfrac{1}{64}=2^x\)

=>\(2^x=\dfrac{1}{2}\cdot\dfrac{1}{2}\cdot...\cdot\dfrac{1}{2}\cdot\dfrac{1}{64}=\left(\dfrac{1}{2}\right)^{30}\cdot\left(\dfrac{1}{2}\right)^6=\dfrac{1}{2^{36}}\)

=>x=-36

a) 1/20 - (x - 8/5) = 1/10

x - 8/5 = 1/20 - 1/10

x - 8/5 = -1/20

x = -1/20 + 8/5

x = 31/20

b) 7/4 - (x + 5/3) = -12/5

x + 5/3 = 7/4 + 12/5

x + 5/3 = 83/20

x = 83/20 - 5/3

x = 149/60

c) x - [17/2 - (-3/7 + 5/3)] = -1/3

x - (17/2 - 26/21) = -1/3

x - 305/42 = -1/3

x = -1/3 + 305/42

x = 97/14

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, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\)\(x\)            = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\) \(x\)           = - \(\dfrac{3}{20}\)

                  \(x\)          =  - \(\dfrac{3}{20}\): \(\dfrac{2}{5}\)

                  \(x\)          = - \(\dfrac{3}{8}\)

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

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

             \(\dfrac{1}{2}\): \(x\) = - \(\dfrac{8}{15}\)

                   \(x\) =  \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))

                    \(x\) =  - \(\dfrac{15}{16}\)

c, - \(\dfrac{2}{3}\): \(x\) + \(\dfrac{5}{8}\) = - \(\dfrac{7}{12}\)

     \(\dfrac{2}{3}\): \(x\)        =  \(\dfrac{7}{12}\) + \(\dfrac{5}{8}\)

     \(\dfrac{2}{3}\) : \(x\)     =  \(\dfrac{29}{24}\)

           \(x\)      = \(\dfrac{2}{3}\) : \(\dfrac{29}{24}\)

            \(x\)     = \(\dfrac{16}{29}\)

\(a,\dfrac{x}{5}=\dfrac{-18}{10}\\ \Rightarrow x=-\dfrac{18}{10}.5\\ \Rightarrow x=-9\\ b,\dfrac{6}{x-1}=\dfrac{-3}{7}\\ \Rightarrow6.7=-3\left(x-1\right)\\ \Rightarrow42=-3x+3\\ \Rightarrow42+3x-3=0\\ \Rightarrow3x+39=0\\ \Rightarrow3x=-39\\ \Rightarrow x=-13\\ c,\dfrac{y-3}{12}=\dfrac{3}{y-3}\\ \Rightarrow\left(y-3\right)^2=36\\ \Rightarrow\left[{}\begin{matrix}y-2=6\\y-2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}y=8\\y=-4\end{matrix}\right.\)

\(d,\dfrac{x}{25}=\dfrac{-5}{x^2}\\ \Rightarrow x^3=-125\\ \Rightarrow x^3=\left(-5\right)^3\\ \Rightarrow x=-5\)

1/2-2y=9/20

=>2y=1/2-9/20=1/20

=>y=1/20:2=1/40

b,3/5:4/3:y=2+7/10=9/20:y=27/10

=>y=9/20:27/10=1/6

c,y+y*3/2-y*1/2=1/10

=>y(1+3/2-1/2)=1/10

=>2y=1/10

=>y=1/10:2=1/20

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)

\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)

\(\Leftrightarrow x=-\dfrac{3}{4}\)

\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)

\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{16}{15}\)

a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

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

b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=-\dfrac{2}{3}\)

c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

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

\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

#YVA6

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32} \)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

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

\(b,\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=\dfrac{-2}{3}\)

\(c,\dfrac{-1}{4}.x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}.x=\dfrac{-1}{20}\)

\(x=\dfrac{-1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

\(d,\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{-1}{4}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

\(#yH\)

\(#NBaoNgoc\)

a) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/3 = y/4 = x/3 + y/4 = 28/7 = 4

=> x = 4 × 3 = 12

=> y = 4 × 4 = 16

Vậy x = 12, y = 16

B) Áp dụng tính chất dãy tỉ số bằng nhau ta được:

X/2 = y/(-5) = x/2 - y/(-5) = (-7)/7 = -1

=> x = -1 × 2 = -2

=> y = -1 × -5 = 5

Vậy x = -2, y = 5

C) làm tương tự như bài a, b

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được: