I- Trắc nghiệm:
1. C
2. D
3. C
4. C
5. D
6. A
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II-Tự luận
Bài 1:
a/ \(2x\left(2x^3+3x+1\right)=4x^4+6x^2+2x\)
b/ \(\left(x+1\right)\left(x-1\right)+\left(x+2\right)^2=x^2-1+x^2+4x+4=2x^2+4x+3\)
Bài 2:
a/ \(3x^3-3x=3x\left(x^2-1\right)=3x\left(x+1\right)\left(x-1\right)\)
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b/ \(x^2-4+\left(x-2\right)^2=\left(x+2\right)\left(x-2\right)+\left(x-2\right)^2=2x\left(x-2\right)\)
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c/ \(12x-x^2-35\)
\(=5x+7x-x^2-35\)
\(=-\left(x^2-5x\right)+\left(7x-35\right)\)
\(=-\left(x\left(x-5\right)\right)+7\left(x-5\right)\)
\(=-\left(x-5\right)\left(x+7\right)\)
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Bài 3: Cho A= 2a2-a+2
\(A=2a^2-a+2\)
\(=2a^2-2a+a-1+3\)
\(=a\left(2a+1\right)-\left(2a+1\right)+3\)
\(=\left(2a+1\right)\left(a-1\right)+3\)
Để A ⋮ 2a+1 => 3 phải chia hết cho 2a+1
\(\left[{}\begin{matrix}2a+1=1\Leftrightarrow2a=0\Leftrightarrow a=0\\2a+1=3\Leftrightarrow2a=2\Leftrightarrow a=1\end{matrix}\right.\)
Vậy: Số nguyên a cần tìm là 0 hoặc 1
