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Bài 1 :
\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)
Bài 2 :
\(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)
=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)
Tick đúng nha
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\(=x\left(2x^2-x-6\right)\)
\(=x\left(2x^2-4x+3x-6\right)\)
\(=x\left[2x\left(x-2\right)+3\left(x-2\right)\right]\)
\(=x\left(x-2\right)\left(2x+3\right)\)
x(2x^2-x-6)
x(2x^2-4x+3x-6)
x[2x(x-2)+3(x-2)]
x(2x+3)(x-2)
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2x2 - 2y2 - 6x - 6x = 2(x2 - y2) - 6(x + y) = 2(x - y)(x + y) -2.3.(x + y) = 2(x + y)[(x - y) - 3] = 2(x + y)(x - y - 3)
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\(=x^4-x^3+3x^3-3x^2+3x^2-3x+9x-9\\ =\left(x-1\right)\left(x^3+3x^2+3x+9\right)\\ =\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)
\(x^4+2x^3+6x-9=x^3\left(x-1\right)+3x^2\left(x-1\right)+3x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+3x+9\right)\)
\(=\left(x-1\right)\left[x^2\left(x+3\right)+3\left(x+3\right)\right]\)
\(=\left(x-1\right)\left(x+3\right)\left(x^2+3\right)\)
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Thợ Đào Mỏ Panda, mày bị điên à, không biết còn trả lời làm cái quái gì
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Ta có x 2 – 6x + 8
= x 2 – 4x – 2x + 8
= x(x – 4) – 2(x – 4)
= (x – 4)(x – 2)
Đáp án cần chọn là: A
\(\text{ 2x² + 6x - 8 }\)
\(2\left(x^2+3x-4\right)=2\left(x-1\right)\left(x+4\right)\)