Đáp án:
\(\begin{array}{l}
B2:\\
A = - 9{x^{16}}{y^{12}}\\
HS: - 9\\
Bậc:28\\
B = - \dfrac{1}{3}{x^6}{y^5}\\
Hs: - \dfrac{1}{3}\\
Bậc:11\\
B4:\\
a)\left[ \begin{array}{l}
x = 0\\
x = 4
\end{array} \right.\\
b)x = 5
\end{array}\)
\(\begin{array}{l}
B3:\\
a)P\left( x \right) + Q\left( x \right) = 4{x^3} + 2{x^2} + 1\\
b)P\left( x \right) - Q\left( x \right) = 2{x^2} - 12x + 13\\
c) - 89
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B2:\\
a)A = \dfrac{1}{9}{x^4}{y^6}.\left( { - 27} \right){x^{12}}{y^6}\\
= - 9{x^{16}}{y^{12}}\\
B = \dfrac{1}{4}.\left( { - \dfrac{1}{2}} \right).\dfrac{8}{3}.{x^{3 + 2 + 1}}.{y^{2 + 1 + 2}}\\
= - \dfrac{1}{3}{x^6}{y^5}\\
b)A = - 9{x^{16}}{y^{12}}\\
HS: - 9\\
Bậc: - 9{x^{16}}{y^{12}}\\
B = - \dfrac{1}{3}{x^6}{y^5}\\
HS: - \dfrac{1}{3}\\
Bậc:6 + 5 = 11\\
B3:\\
a)P\left( x \right) + Q\left( x \right) = 2{x^3} + 10{x^2} - 6x + 7 + 2{x^3} - 8{x^2} + 6x - 6\\
= 4{x^3} + 2{x^2} + 1\\
b)P\left( x \right) - Q\left( x \right) = 2{x^3} + 10{x^2} - 6x + 7 - 2{x^3} + 8{x^2} - 6x + 6\\
= 2{x^2} - 12x + 13\\
c)P\left( x \right) + Q\left( x \right) = 4{x^3} + 2{x^2} + 1\\
Thay:x = - 3\\
\to 4.{\left( { - 3} \right)^3} + 2.{\left( { - 3} \right)^2} + 1 = - 89\\
B4:\\
a)A\left( x \right) = 0\\
\to x\left( {6 - x} \right) = 0\\
\to \left[ \begin{array}{l}
x = 0\\
x = 6
\end{array} \right.\\
b)B\left( x \right) = 0\\
\to \dfrac{1}{3}x - \dfrac{5}{3} = 0\\
\to x = 5
\end{array}\)
