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\(\dfrac{x^5+x^4-15x^3-5x^2+34x+24}{x^2+5x+4}\)
\(=\dfrac{x^5+5x^4+4x^3-4x^4-20x^3-16x^2+x^3+5x^2+4x+6x^2+30x+24}{x^2+5x+4}\)
\(=x^3-4x^2+x+6\)
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a) \(12x^3+8x^2-3x-2=4x^2\left(3x+2\right)-\left(3x+2\right)\)
\(=\left(3x+2\right)\left(4x^2-1\right)=\left(3x+2\right)\left(2x-1\right)\left(2x+1\right)\)
b) \(18x^3+27x^2-2x-3=9x^2\left(2x+3\right)-\left(2x+3\right)\)
\(=\left(2x+3\right)\left(9x^2-1\right)=\left(2x+3\right)\left(3x-1\right)\left(3x+1\right)\)
c) \(8x^3+4x^2-34x+15=4x^2\left(2x-3\right)+8x\left(2x-3\right)-5\left(2x-3\right)\)
\(=\left(2x-3\right)\left(4x^2+8x-5\right)=\left(2x-3\right)\left(2x-1\right)\left(2x+5\right)\)
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\(15x^2-34x+15\)
\(=15x^2-25x-9x+15\)
\(=5x\left(3x-5\right)-3\left(3x-5\right)\)
\(=\left(5x-3\right)\left(3x-5\right)\)
15x4-34x4
=x4(15-34)
=x4.(-19)
Phải hok ạ ???