Đáp án:
$\begin{array}{l}
B1)\\
a){\left( { - 2{x^2}y} \right)^2}.{\left( {3x{y^2}z} \right)^2}\\
= 4{x^4}{y^2}.9{x^2}{y^4}{z^2}\\
= 36{x^6}{y^6}{z^2}
\end{array}$
b) Bậc của tích là: 6+6+2=14
Hệ số là: 36
Biến là: ${x^6}{y^6}{z^2}$
Bài 2:
$\begin{array}{l}
a)P\left( x \right) = 5{x^5} + 3x - 4{x^4} - 2{x^3} + 6 + 4{x^2}\\
= 5{x^5} - 4{x^4} - 2{x^3} + 4{x^2} + 3x + 6\\
Q\left( x \right) = 2{x^4} - x + 3{x^2} - 2{x^3} + \dfrac{1}{4} - {x^5}\\
= - {x^5} + 2{x^4} - 2{x^3} + 3{x^2} - x + \dfrac{1}{4}\\
b)\\
P\left( x \right) + Q\left( x \right)\\
= 5{x^5} - 4{x^4} - 2{x^3} + 4{x^2} + 3x + 6\\
- {x^5} + 2{x^4} - 2{x^3} + 3{x^2} - x + \dfrac{1}{4}\\
= 4{x^5} - 2{x^4} - 4{x^3} + 7{x^2} + 2x + \dfrac{{25}}{4}\\
P\left( x \right) - Q\left( x \right)\\
= 5{x^5} - 4{x^4} - 2{x^3} + 4{x^2} + 3x + 6\\
- \left( { - {x^5} + 2{x^4} - 2{x^3} + 3{x^2} - x + \dfrac{1}{4}} \right)\\
= 6{x^5} - 6{x^4} + {x^2} + 4x + \dfrac{{23}}{4}\\
c)Khi:\left| x \right| = 1 \Rightarrow \left[ \begin{array}{l}
x = 1\\
x = - 1
\end{array} \right.\\
+ Khi:x = 1\\
\Rightarrow P\left( x \right) + Q\left( x \right)\\
= 4{x^5} - 2{x^4} - 4{x^3} + 7{x^2} + 2x + \dfrac{{25}}{4}\\
= 4.1 - 2.1 - 4.1 + 7.1 + 2.1 + \dfrac{{25}}{4}\\
= \dfrac{{53}}{4}\\
+ Khi:x = - 1\\
\Rightarrow P\left( x \right) + Q\left( x \right)\\
= 4{x^5} - 2{x^4} - 4{x^3} + 7{x^2} + 2x + \dfrac{{25}}{4}\\
= 4.\left( { - 1} \right) - 2.1 - 4.\left( { - 1} \right) + 7.1 + 2.\left( { - 1} \right) + \dfrac{{25}}{4}\\
= \dfrac{{37}}{4}
\end{array}$
