a) (x+2) ³-(x-3)(x+3)+10
b)-3x(-4x-1)+6x(4-2x)-5
2. Tìm m để đa thức 2x ³+3x ²-x+m chia hết cho đa thức 2x+1
Bài 2:
a) 2x ²-6x
b) 3x(5y-2)-10(2-5y)
c) x mũ 4+x ³+2x ²+x+1
a) (x+2) ³-(x-3)(x+3)+10
b)-3x(-4x-1)+6x(4-2x)-5
2. Tìm m để đa thức 2x ³+3x ²-x+m chia hết cho đa thức 2x+1
Bài 2:
a) 2x ²-6x
b) 3x(5y-2)-10(2-5y)
c) x mũ 4+x ³+2x ²+x+1
Câu hỏi mới
a){x^3} + 5{x^2} + 12x + 27\\
b)27x – 5\\
2)\,m = – 1\\
2)a)2x\left( {x – 3} \right)\\
b)\left( {5y – 2} \right)\left( {3x + 10} \right)\\
c)\left( {{x^2} + x + 1} \right)\left( {{x^2} + 1} \right)
\end{array}$
a){\left( {x + 2} \right)^3} – \left( {x – 3} \right)\left( {x + 3} \right) + 10\\
= {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} – \left( {{x^2} – {3^2}} \right) + 10\\
= {x^3} + 6{x^2} + 12x + 8 – {x^2} + 9 + 10\\
= {x^3} + 5{x^2} + 12x + 27\\
b) – 3x\left( { – 4x – 1} \right) + 6x\left( {4 – 2x} \right) – 5\\
= 12{x^2} + 3x + 24x – 12{x^2} – 5\\
= 27x – 5\\
2)\\
2{x^3} + 3{x^2} – x + m\\
= 2{x^3} + {x^2} + 2{x^2} + x – 2x – 1 + 1 + m\\
= {x^2}\left( {2x + 1} \right) + x\left( {2x + 1} \right) – \left( {2x + 1} \right) + m + 1\\
= \left( {2x + 1} \right)\left( {{x^2} + x – 1} \right) + m + 1\\
Khi:\left( {2{x^3} + 3{x^2} – x + m} \right) \vdots \left( {2x + 1} \right)\\
\Leftrightarrow m + 1 = 0\\
\Leftrightarrow m = – 1\\
Vậy\,m = – 1\\
2)a)2{x^2} – 6x = 2x\left( {x – 3} \right)\\
b)3x\left( {5y – 2} \right) – 10\left( {2 – 5y} \right)\\
= 3x\left( {5y – 2} \right) + 10\left( {5y – 2} \right)\\
= \left( {5y – 2} \right)\left( {3x + 10} \right)\\
c){x^4} + {x^3} + 2{x^2} + x + 1\\
= {x^4} + {x^3} + {x^2} + {x^2} + x + 1\\
= {x^2}\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
= \left( {{x^2} + x + 1} \right)\left( {{x^2} + 1} \right)
\end{array}$