Bài 1 trang 5 SGK Toán lớp 8 tập 1

\(a)\ x^2(5x^3-x-\frac{1}{2})=x^2.5x^3+x^2.(-x)+x^2.(-\frac{1}{2})\)

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

\(b)\ (3xy-x^2+y)\ \frac{2}{3x^2y}=\frac{2}{3x^2y}.3xy+\frac{2}{3x^2y}.(-x^2)+\frac{2}{3x^2y}.y\)

\(=2x^3y^2-\frac{2}{3x^4y}+\frac{2}{3x^2y^2}\)

\(c)\ (4x^3-5xy+2x)(-\frac{1}{2xy})=-\frac{1}{2xy}.4x^3+(-\frac{1}{2xy}).(-5xy)+(-\frac{1}{2xy}).2x\)

\(=-2x^4y+\frac{5}{2x^2y^2}-x^2y.\)

Lời giải chi tiết

a. \({x^2}.\left( {5{x^3} - x - \frac{1}{2}} \right)\)

\(\begin{matrix} = x.5{x^3} - {x^2}.x - {x^2}.\dfrac{1}{2} \hfill \\ = 5{x^4} - {x^3} - \dfrac{1}{2}.{x^2} \hfill \\ \end{matrix}\)

b. \(\left( {3xy - {x^2} + y} \right).\frac{2}{3}{x^2}y\)

\(\begin{matrix} = 3xy.\dfrac{2}{3}.{x^2}y - {x^2}.\dfrac{2}{3}{x^2}y + y.\dfrac{2}{3}{x^2}y \hfill \\ = 2{x^3}{y^2} - \dfrac{2}{3}{x^4}y + \dfrac{2}{3}{x^2}{y^2} \hfill \\ \end{matrix}\)

c. \(\left( {4{x^3} - 5xy + 2x} \right)\left( { - \frac{1}{2}xy} \right)\)

\(\begin{matrix} = 4{x^3}.\left( { - \dfrac{1}{2}xy} \right) - 5xy.\left( { - \dfrac{1}{2}xy} \right) + 2x.\left( { - \dfrac{1}{2}xy} \right) \hfill \\ = - 2{x^4}y + \dfrac{5}{2}{x^2}{y^2} - {x^2}y \hfill \\ \end{matrix}\)

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