giúp mình làm với toán

Đáp án:

$\begin{array}{l} 1,\\ a,\\ 15{x^3} - 6{x^2} - 39x\\ b,\\  - {x^3}y - 3{x^2}{y^3} + 3xy\\ c,\\ {x^5}y - \dfrac{1}{5}{x^3}{y^3} - \dfrac{1}{2}{x^2}y\\ 2,\\ a,\\  - 3{x^3} - 3x\\ b,\\  - 11x + 24\\ c,\\ 3{x^4} - \dfrac{5}{2}{x^3} + 2\\ 3,\\ a,\\ x =  - 2\\ b,\\ x = 3\\ c,\\ Phương\,\,trình\,\,vô\,\,nghiệm \end{array}$

Giải thích các bước giải:

 Ta có:

$\begin{array}{l} 1,\\ a,\\ 3x\left( {5{x^2} - 2x - 13} \right)\\  = 3x.5{x^2} - 3x.2x - 3x.13\\  = 15{x^3} - 6{x^2} - 39x\\ b,\\ \left( {{x^2} + 3xy - 3} \right).\left( { - xy} \right)\\  = {x^2}.\left( { - xy} \right) + 3xy.\left( { - xy} \right) - 3.\left( { - xy} \right)\\  =  - {x^3}y - 3{x^2}{y^3} + 3xy\\ c,\\ \dfrac{1}{2}{x^2}y\left( {2{x^3} - \dfrac{2}{5}x{y^2} - 1} \right)\\  = \dfrac{1}{2}{x^2}y.2{x^3} - \dfrac{1}{2}{x^2}y.\dfrac{2}{5}x{y^2} - \dfrac{1}{2}{x^2}y.1\\  = {x^5}y - \dfrac{1}{5}{x^3}{y^3} - \dfrac{1}{2}{x^2}y\\ 2,\\ a,\\ x\left( {2{x^2} - 3} \right) - {x^2}\left( {5x + 1} \right) + {x^2}\\  = x.2{x^2} - 3.x - {x^2}.5x - {x^2}.1 + {x^2}\\  = 2{x^3} - 3x - 5{x^3} - {x^2} + {x^2}\\  =  - 3{x^3} - 3x\\ b,\\ 3x\left( {x - 2} \right) - 5x\left( {1 - x} \right) - 8.\left( {{x^2} - 3} \right)\\  = 3x.x - 3x.2 - 5x.1 + 5x.x - 8{x^2} + 8.3\\  = 3{x^2} - 6x - 5x + 5{x^2} - 8{x^2} + 24\\  = \left( {3{x^2} + 5{x^2} - 8{x^2}} \right) + \left( { - 6x - 5x} \right) + 24\\  =  - 11x + 24\\ c,\\ \dfrac{1}{2}{x^3}\left( {6x - 3} \right) - x.\left( {{x^2} + \dfrac{1}{2}} \right) + \dfrac{1}{2}\left( {x + 4} \right)\\  = \dfrac{1}{2}{x^3}.6x - \dfrac{1}{2}{x^3}.3 - x.{x^2} - x.\dfrac{1}{2} + \dfrac{1}{2}.x + \dfrac{1}{2}.4\\  = 3{x^4} - \dfrac{3}{2}{x^3} - {x^3} - \dfrac{1}{2}x + \dfrac{1}{2}x + 2\\  = 3{x^4} - \dfrac{5}{2}{x^3} + 2\\ 3,\\ a,\\ 2x\left( {x - 5} \right) - x\left( {3 + 2x} \right) = 26\\  \Leftrightarrow 2x.x - 2x.5 - x.3 - x.2x = 26\\  \Leftrightarrow 2{x^2} - 10x - 3x - 2{x^2} = 26\\  \Leftrightarrow  - 13x = 26\\  \Leftrightarrow x =  - 2\\ b,\\ 2x\left( {{x^2} - 2} \right) + {x^2}\left( {1 - 2x} \right) - {x^2} =  - 12\\  \Leftrightarrow 2x.{x^2} - 2x.2 + {x^2} - {x^2}.2x - {x^2} =  - 12\\  \Leftrightarrow 2{x^3} - 4x + {x^2} - 2{x^3} - {x^2} =  - 12\\  \Leftrightarrow  - 4x =  - 12\\  \Leftrightarrow x = 3\\ c,\\ 3x\left( {x + 1} \right) - 2x\left( {x + 2} \right) =  - 1 - x\\  \Leftrightarrow 3x.x + 3x.1 - 2x.x - 2x.2 + 1 + x = 0\\  \Leftrightarrow 3{x^2} + 3x - 2{x^2} - 4x + 1 + x = 0\\  \Leftrightarrow \left( {3{x^2} - 2{x^2}} \right) + \left( {3x - 4x + x} \right) + 1 = 0\\  \Leftrightarrow {x^2} + 1 = 0\\  \Leftrightarrow {x^2} =  - 1\\ {x^2} \ge 0,\,\,\,\forall x \Rightarrow Phương\,\,trình\,\,vô\,\,nghiệm \end{array}$