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Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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\(3x^3-6x-3x^2\)
\(=3x\left(x^2-x-2\right)\)
\(=3x\left[\left(x^2-2x\right)+\left(x-2\right)\right]\)
\(=3x\left[x.\left(x-2\right)+\left(x-2\right)\right]\)
\(=3x\left(x-2\right)\left(x+1\right)\)
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x3 - 2x2 + 6x - 5 = x3 - x2 - x2 + x + 5x - 5 = x2(x - 1) - x(x - 1) + 5(x - 1) = (x2 - x + 5)(x - 1)
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nếu có thể các bạn dùng phương pháp đồng nhất hệ số hộ mình nhé ^^
\(x^4-3x^3+6x^2-5x+3\)
\(=x^4-2x^3+3x^2-x^3+2x^2-3x+x^2-2x+3\)
\(=x^2\left(x^2-2x+3\right)-x\left(x^2-2x+3\right)+\left(x^2-2x+3\right)\)
\(=\left(x^2-x+1\right)\left(x^2-2x+3\right)\)
Đây là phương pháp hệ số bất định. Chắc bạn đang học nâng cao nên cũng đọc rồi.
Chúc bạn học tốt.
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a)\(x^5-x=x\left(x^4-1\right)=x\left(x^2-1\right)\left(x^2+1\right)=x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
b)\(3x^2+5x-2=3x^2+6x-x-2=3x\left(x+2\right)-\left(x+2\right)=\left(3x-1\right)\left(x+2\right)\)
c)\(4x^3+14x^2+6x=2x\left(2x^2+7x+3\right)=2x\left(2x^2+6x+x+3\right)\)
\(=2x\left[2x\left(x+3\right)+\left(x+3\right)\right]=2x\left(x+3\right)\left(2x+1\right)\)
\(x^5-x=x\left(x^4-1\right)\)
\(=x\left(x^2-1\right)\left(x^2+1\right)\)
\(=x\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
Answer:
\((3x-2)^2(6x-5)(6x-3)-5\)
\(=(9x^2-12x+4)(6x-5)(6x-3)-5\)
\(=(54x^3-117x^2+84x-20)(6x-3)-5\)
\(=324x^4-864x^3+855x^2-372x+60-5\)
\(=324x^3-864x^3+855x^2-372x+55\)
\(=(9x^2-12x+5)(36x^2-48x+11)\)