\(\left(x+\dfrac{1}{3}\right)^3:3=-\dfrac{1}{81}\\ \Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=-\dfrac{1}{27}\\ \Leftrightarrow x+\dfrac{1}{2}=-\dfrac{1}{3}\\ \Leftrightarrow x=-\dfrac{5}{6}\)

\(\left(\dfrac{1}{3}\right)^{2x-1}=3^5\\ \Leftrightarrow3^{1-2x}=3^5\\ \Leftrightarrow1-2x=5\\ \Leftrightarrow2x=-4\\ \Leftrightarrow x=-2.\)

\(x\left(x^2-y\right)-x^2\left(x+y\right)+xy\left(x-1\right)=x^3-xy-x^3-x^2y+x^2y-xy=-xy\).

a, \(5x^2y\cdot2xy^2=10x^3y^3\)

b, \(\dfrac{3}{4}xy\cdot8x^3y^2=6x^4y^3\)

c, \(1,5xy^2z^3\cdot2x^3y^2z=3x^3y^4z^4\)

\(\left(2x+3y\right)\left(x^2-5xy+4y^2\right)=2x^3-10x^2y+8xy^2+3x^2y-15xy^2+12y^3=2x^3+12y^3-7x^2y-7xy^2.\)

a. \(A+B=2x^2y+3xyz-2x+5+3xyz-2x^2y+x-4=6xyz-x+1\\ A-B=2x^2y+3xyz-2x+5-3xyz+2x^2y-x+4=4x^2y-3x+9\)

b. Khi x = 0,5, y = -2, z = 1, ta có: 

\(A=2\cdot0,5^2\cdot\left(-2\right)+3\cdot0,5\cdot\left(-2\right)\cdot1-2\cdot0,5+5=0\\ A+B=6\cdot0,5\cdot\left(-2\right)\cdot1-0,5+1=-\dfrac{11}{2}\)

\(\left(2x-5\right)^2=\left(x-\dfrac{5}{2}\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=x-\dfrac{5}{2}\\2x-5=\dfrac{5}{2}-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\3x=\dfrac{15}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{5}{2}\)

Vậy x = \(\dfrac{5}{2}\)