Đáp án:
$\begin{array}{l}
b1)\\
a)A = \frac{3}{4}.x{y^2}.8{x^2}{y^3}\\
= \frac{3}{4}.8.x.{x^2}.{y^2}.{y^3}\\
= 6{x^3}.{y^5}
\end{array}$
b) Hệ số: 6
Phần biến:${x^3}{y^5}$
Bậc: 8
$\begin{array}{l}
c)Khi:x = 2;y = - 1\\
\Leftrightarrow A = {6.2^3}.{\left( { - 1} \right)^5} = 6.8.\left( { - 1} \right) = - 48
\end{array}$
$\begin{array}{l}
B2)\\
a)A\left( x \right) = 7{x^3} + 3{x^4} + 6 - 4{x^2} + 2x\\
= 3{x^4} + 7{x^3} - 4{x^2} + 2x + 6\\
B\left( x \right) = 5x - 2{x^3} + 8{x^4} + {x^2} - 1\\
= 8{x^4} - 2{x^3} + {x^2} + 5x - 1\\
b)A\left( x \right) + B\left( x \right)\\
= 3{x^4} + 7{x^3} - 4{x^2} + 2x + 6\\
+ 8{x^4} - 2{x^3} + {x^2} + 5x - 1\\
= 11{x^4} + 5{x^3} - 3{x^2} + 7x + 5\\
c)A\left( x \right) - B\left( x \right)\\
= 3{x^4} + 7{x^3} - 4{x^2} + 2x + 6\\
- \left( {8{x^4} - 2{x^3} + {x^2} + 5x - 1} \right)\\
= 3{x^4} + 7{x^3} - 4{x^2} + 2x + 6\\
- 8{x^4} + 2{x^3} - {x^2} - 5x + 1\\
= - 5{x^4} + 9{x^3} - 5{x^2} - 3x + 7\\
B3)\\
a)A\left( x \right) = 0\\
\Leftrightarrow 7x + 28 = 0\\
\Leftrightarrow 7x = - 28\\
\Leftrightarrow x = - 4\\
Vậy\,x = - 4\\
b)B\left( x \right) = 0\\
\Leftrightarrow 49 - {x^2} = 0\\
\Leftrightarrow {x^2} = 49\\
\Leftrightarrow x = 7;x = - 7\\
Vậy\,x = 7;x = - 7\\
c)C\left( x \right) = 0\\
\Leftrightarrow {x^3} + 81x = 0\\
\Leftrightarrow x\left( {{x^2} + 81} \right) = 0\\
\Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$
