Phân tích đa thức sau thành nhân tử:(x^2-x-2)(x^2+9x+18)-28
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(x2 + x + 2)(x2 + 9x + 18) - 28
= x4 + 10x3 + 29x2 + 36x + 36 - 28
= x4 + 10x3 + 29x2 + 36x + 8
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\(x^3-9x^2+x=x\left(x^2-9x+1\right)\)
\(x^3+13x^2+x=x\left(x^2+13x+1\right)\)
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\(\Rightarrow\left(x^2+2x\right)^2+9\left(x^2+2x\right)+20=\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\cdot\left(x^2+2x\right)+20=\left(x^2+2x\right)\left(x^2+2x+4\right)+5\left(x^2+2x+4\right)=\left(x^2+2x+4\right)\left(x^2+2x+5\right)\)
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\(\left(x^2+2x\right)^2+4\left(x^2+2x\right)+5\left(x^2+2x\right)+20\)
\(=\left(x^2+2x\right)\left(x^2+2x+4\right)+5\left(x^2+2x+4\right)\)
\(=\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)
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Bài 1:
\(=\left(3x-1\right)^2-9y^2\)
=(3x-1-3y)(3x-1+3y)
=(3x−1)2−9y2=(3x−1)2−9y2
=(3x-1-3y)(3x-1+3y)
Tham khảo ạ
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Ta có: \(x^2-3x-28\)
\(=x^2-7x+4x-28\)
\(=x\left(x-7\right)+4\left(x-7\right)\)
\(=\left(x-7\right)\left(x+4\right)\)