Đáp án:
B2:
\(b)B = - x{y^3} - \dfrac{2}{5}{x^2}y + \dfrac{{31}}{{30}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)M = - 4{x^3} + 5{x^4} - 7 - 5{x^4} + 4{x^3} + 2x - \dfrac{1}{2}\\
= 2x - \dfrac{{15}}{2}\\
b)M = N\\
\to 2x - \dfrac{{15}}{2} = 6x - 15\\
\to 4x - 15 = 12x - 30\\
\to 8x = 15\\
\to x = \dfrac{{15}}{8}\\
2)A.B = - \dfrac{2}{3}{b^3}{c^2}.\left( { - {a^3}b{c^2}} \right).\dfrac{1}{4}{c^3}.\left( { - b} \right).b{c^2}\\
= \dfrac{2}{3}{a^3}{b^4}{c^2}.\left( { - \dfrac{1}{4}} \right){b^2}{c^5}\\
= - \dfrac{1}{6}{a^3}{b^6}{c^7}\\
Bậc:3 + 6 + 7 = 16\\
B2:\\
a)A = \dfrac{1}{2}x{y^3} + 3{x^2}y - \dfrac{4}{5} - x{y^3} - \dfrac{{12}}{5}{x^2}y + \dfrac{1}{3}\\
= - \dfrac{2}{3}x{y^3} + \dfrac{3}{5}{x^2}y - \dfrac{7}{{15}}\\
b)B = A - {x^2}y + \dfrac{3}{2} - \dfrac{1}{3}x{y^3}\\
= - \dfrac{2}{3}x{y^3} + \dfrac{3}{5}{x^2}y - \dfrac{7}{{15}} - {x^2}y + \dfrac{3}{2} - \dfrac{1}{3}x{y^3}\\
= - x{y^3} - \dfrac{2}{5}{x^2}y + \dfrac{{31}}{{30}}
\end{array}\)
