Đáp án:
\(\begin{array}{l}
1,\\
a,18{x^2}{y^4} - 3{x^2}{y^3} + \dfrac{9}{2}{x^2}{y^2}\\
b, - 2{x^3}y + \dfrac{3}{8}{x^2}{y^2} + 3\\
2,\\
a,x = - \dfrac{{20}}{7}\\
b,x = 4\\
3,\\
a,15{x^4} - 21{x^3} + 6{x^2}\\
b,\dfrac{4}{3}{x^3}{y^2} - \dfrac{8}{3}{x^2}{y^2} - \dfrac{2}{3}x{y^3}\\
c,2{x^3} + 2{x^2}y - 12{x^2} - 3x
\end{array}\)
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
a,\\
6x{y^2}.\left( {3x{y^2} - \dfrac{1}{2}xy + \dfrac{3}{4}x} \right)\\
= 6x{y^2}.3x{y^2} - 6x{y^2}.\dfrac{1}{2}xy + 6x{y^2}.\dfrac{3}{4}x\\
= 18{x^2}{y^4} - 3{x^2}{y^3} + \dfrac{9}{2}{x^2}{y^2}\\
b,\\
\left( { - \dfrac{1}{2}x} \right).\left( {4{x^2}y - \dfrac{3}{4}x{y^2} - 6} \right)\\
= \left( { - \dfrac{1}{2}x} \right).4{x^2}y - \left( { - \dfrac{1}{2}x} \right).\dfrac{3}{4}x{y^2} - \left( { - \dfrac{1}{2}} \right).6\\
= - 2{x^3}y + \dfrac{3}{8}{x^2}{y^2} + 3\\
2,\\
a,\\
3x\left( {x + 1} \right) - x.\left( {3x - 4} \right) = - 20\\
\Leftrightarrow \left( {3{x^2} + 3x} \right) - \left( {3{x^2} - 4x} \right) = - 20\\
\Leftrightarrow 3{x^2} + 3x - 3{x^2} + 4x = - 20\\
\Leftrightarrow 7x = - 20\\
\Leftrightarrow x = - \dfrac{{20}}{7}\\
b,\\
x\left( {x + 2} \right) - 2x\left( {\dfrac{1}{2}x - 1} \right) = 16\\
\Leftrightarrow \left( {{x^2} + 2x} \right) - \left( {2x.\dfrac{1}{2}x - 2x.1} \right) = 16\\
\Leftrightarrow \left( {{x^2} + 2x} \right) - \left( {{x^2} - 2x} \right) = 16\\
\Leftrightarrow {x^2} + 2x - {x^2} + 2x = 16\\
\Leftrightarrow 4x = 16\\
\Leftrightarrow x = 16:4\\
\Leftrightarrow x = 4\\
3,\\
a,\\
3{x^2}.\left( {5{x^2} - 7x + 2} \right)\\
= 3{x^2}.5{x^2} - 3{x^2}.7x + 3{x^2}.2\\
= 15{x^4} - 21{x^3} + 6{x^2}\\
b,\\
\dfrac{2}{3}xy\left( {2{x^2}y - 4xy - {y^2}} \right)\\
= \dfrac{2}{3}xy.2{x^2}y - \dfrac{2}{3}xy.4xy - \dfrac{2}{3}xy.{y^2}\\
= \dfrac{4}{3}{x^3}{y^2} - \dfrac{8}{3}{x^2}{y^2} - \dfrac{2}{3}x{y^3}\\
c,\\
2{x^2}\left( {x + y} \right) - 3x\left( {2x + 1} \right) - 6{x^2}\\
= 2{x^2}.x + 2{x^2}.y - \left( {3x.2x + 3x.1} \right) - 6{x^2}\\
= 2{x^3} + 2{x^2}y - \left( {6{x^2} + 3x} \right) - 6{x^2}\\
= 2{x^3} + 2{x^2}y - 6{x^2} - 3x - 6{x^2}\\
= 2{x^3} + 2{x^2}y - 12{x^2} - 3x
\end{array}\)
