Ta có: \(\left(\dfrac{x+2}{3x}+\dfrac{2}{x+1}-3\right):\dfrac{2-4x}{x+1}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\left(\dfrac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\dfrac{6x}{3x\left(x+1\right)}-\dfrac{9x\left(x+1\right)}{3x\left(x+1\right)}\right):\dfrac{2-4x}{x+1}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\dfrac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\dfrac{-8x^2+2}{3x\left(x+1\right)}\cdot\dfrac{x+1}{2-4x}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\dfrac{-2\left(4x^2-1\right)}{3x\cdot2\cdot\left(1-2x\right)}\cdot\dfrac{3x}{x^2-3x-1}\)

\(=\dfrac{2\left(1-2x\right)\left(2x+3\right)}{6x\left(1-2x\right)}\cdot\dfrac{3x}{x^2-3x-1}\)

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

a) = x^2 - 9 - (x^2 + 3x - 10)

= -3x + 1

b) = 3x + 1 - 3x + 19

= 20

a: \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\)

\(=x^2-9-x^2-3x+10\)

\(=-3x+1\)

b: \(\dfrac{27x^3+1}{9x^2-3x+1}-\left(3x-19\right)\)

\(=3x+1-3x+19\)

=20

a) \(5x^2\left(3x^2-7x+2\right)\)

\(=5x^2.3x^2+5x^2.\left(-7x\right)+5x^2.2\)

\(=\left(5.3\right)\left(x^2.x^2\right)+\left[5.\left(-7\right)\right].\left(x^2.x\right)+\left(5.2\right).x^2\)

\(=15x^{2+2}+\left(-35\right).x^{2+1}+10.x^2\)

\(=15x^4-35x^3+10x^2\)

\(=\dfrac{2x+6}{x\left(3x-1\right)}+\dfrac{x+3}{3x-1}\)

\(=\dfrac{2x+6+x^2+3x}{x\left(3x-1\right)}\)

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

\(a,\left(3x+1\right)\left(3x-1\right)-\left(18x^3+5x^2-2x\right):2x\\ =\left(9x^2-1\right)-\left(9x^2+\dfrac{5}{2}x-1\right)\\ =9x^2-1-9x^2-\dfrac{5}{2}x+1=\dfrac{5}{2}x\)

\(b,3x\left(x-2021\right)-x+2021=0\\ \Rightarrow b,3x\left(x-2021\right)-\left(x-2021\right)=0\\ \Rightarrow\left(x-2021\right)\left(3x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2021\\x=\dfrac{1}{3}\end{matrix}\right.\)

a: \(=\dfrac{x+1+3}{x-3}=\dfrac{x+4}{x-3}\)

\(a,=12x^2-4x-6x-2-x-3=12x^2-11x-5\\ b,=12x^2-9x-12x^2-4x+5=5-13x\\ c,=12x^3-4x^2-12x^3-12x^2+7x-3=-16x^2+7x-3\\ d,=\left(x^2-4\right)\left(x^2+4\right)=x^4-16\)

\(\begin{array}{l}(3x + 1)({x^2} - 2x + 1)\\ = 3x({x^2} - 2x + 1) + 1({x^2} - 2x + 1)\\ = 3{x^3} - 6{x^2} + 3x + {x^2} - 2x + 1\\ = 3{x^3} - 5{x^2} + x + 1\end{array}\)

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

\( \Rightarrow (3{x^3} - 5{x^2} + x + 1):(3x + 1) = {x^2} - 2x + 1\) 

bảo người ta ngu thì mày làm đi

\(a,=27x^3+27x^2+9x+1\)

\(b,=\dfrac{x^3}{27}-\dfrac{x^2}{3}+x-1\)

\(c,=-\left(27x^3-27x^2y^2+9xy^4-y^6\right)\)

\(=-27x^3+27x^2y^2-9xy^4+y^6\)

\(d,=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)

a) \(\left(3x+1\right)^3=27x^3+27x^2+9x+1\)

b) \(\left(\dfrac{x}{3}-1\right)^3=\dfrac{x^3}{27}-\dfrac{x^2}{3}\)

c) \(\left(-y^2+3x\right)^3=27x^3-27x^2y^2+9xy^4-y^6\)

d) \(\left(\dfrac{x}{y}-\dfrac{2y}{x}\right)^3=\dfrac{x^3}{y^3}-\dfrac{6x}{y}+\dfrac{12y}{x}-\dfrac{8y^3}{x^3}\)