\(a,x:\left(\dfrac{-1}{5}\right)^2=\left(\dfrac{-1}{5}\right)\)

\(b,\left(\dfrac{3}{5}\right)^5:x=\left(\dfrac{3}{5}\right)^2\)

\(c,\left(\dfrac{2}{5}\right)^7:x=\left(\dfrac{2}{5}\right)^3\)

\(d,\left(\dfrac{1}{3}\right)^5:x=\left(\dfrac{1}{3}\right)^3\)

\(Bài1:Tính\)

\(a,\left(\dfrac{-2}{3}\right)^4\)

\(b,\left(-3\dfrac{1}{3}\right)^2\)

\(c,\left(-1\dfrac{5}{7}\right)^2\)

\(d,\left(-1\dfrac{3}{4}\right)^3\)

\(e,\left(-0,4\right)^2\)

20 - 2 ( x + 4 ) = 4

       2 ( x + 4 ) = 20 - 4

       2 ( x + 4 ) = 16

            x + 4    = 16 : 2

            x + 4    = 8

             x          = 8 - 4

             x          = 4

Vậy x - 4

\(a,\dfrac{-2}{7}.X=\dfrac{-3}{14}\)

\(b,\dfrac{-5}{2}:X=\dfrac{10}{3}\)

\(c,\dfrac{4}{3}+\dfrac{5}{8}:X=\dfrac{1}{12}\)

\(d,\dfrac{2}{3}+\dfrac{7}{4}:X=\dfrac{5}{6}\)

\(e,\left(x-\dfrac{1}{3}\right).\left(x+\dfrac{2}{5}\right)=0\)

\(f,\left(3x-1\right).\left(2-5x\right)=0\)

\(g,\left(x-\dfrac{5}{3}\right).\left(2x+1\right)=0\)

\(h,x.\left(5x+2\right)=0\)