a) \(x=-\dfrac{3}{5}\times\dfrac{9}{7}=-\dfrac{27}{35}\)

b) \(x\left(0,4-\dfrac{1}{5}\right)=\dfrac{3}{4}\)

\(x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)

a, \(x=-3,5.\dfrac{9}{7}=-\dfrac{9}{2}\)

b, \(\dfrac{2}{5}x-\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)

\(36.4-\left(86-7.12\right)^2:4-2021^0\)

\(=144-2^2:4-1\)

\(=144-4:4-1\)

\(=144-1-1\)

\(=142\)

2021^0=0 chứ ạ

\(-\frac{3}{6}=\frac{x-1}{6}\)

\(\Rightarrow-3=x-1\)

\(\Rightarrow x=-3+1\)

\(\Rightarrow x=-2\)

-1/2=x-1/6

=> (-1).6=(x-1).2        (Áp dụng tích chất 2 phân số bằng nhau thì 2 tích chéo bằng nhau)

=> (x-1).2=-6

=> x-1     =(-6):2

=> x-1     =-3

=> x        =(-3)+1

=> x        =-2

Vậy x=-2

\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}\right)^2\cdot\left(-1\right)^5}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)

\(=\frac{\frac{2^3}{3^3}\cdot\frac{3^2}{4^2}\cdot\left(-1\right)}{\frac{2^2}{5^2}\cdot\frac{\left(-5\right)^3}{12^3}}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\frac{-125}{1728}}=\frac{-\frac{1}{6}}{-\frac{5}{432}}=\left(-\frac{1}{6}\right)\cdot\left(-\frac{432}{5}\right)=\frac{72}{5}\)

Ta có: ( x - 2) x ( y + 3) = -13 = (-13) x 1 = (-1) x 13

* Nếu x - 2 = -13 => x = (-13) + 2 = -11

         y + 3 = 1 => y = 1-3 = -2 

* Nếu x-2 = -1 => x = (-1) + 2 = 1

         y + 3 = 13 => y = 13 - 3 = 10

Vậy có 2 cặp x;y x;y(-11;-2)

                          x;y(1;10)

 Để  60+x không chia hết cho 6

→   x   không chia hết cho 6 (  do 60 chia hết cho 6  )

→  x   ∈   {22;45} 

Vậy x ∈  {22;45}