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NT
11 tháng 6 2023 lúc 8:54

a: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=2x^2-3x+1\)

b: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)

d: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=\left(x-1\right)^2\)

h: \(=\dfrac{x^3+x^2-3x^2-3x+8x+8}{x+1}=x^2-3x+8\)

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NT
11 tháng 6 2023 lúc 10:37

\(c,\)

\(\left(6x^3-19x^2+23x-12\right):\left(2x-3\right)\)

\(=\left(6x^3-10x^2-9x^2+8x+15x-12\right):\left(2x-3\right)\)

\(=\left[\left(6x^3-10x^2+8x\right)-\left(9x^2-15x+12\right)\right]:\left(2x-3\right)\)

\(=\left[2x\left(3x^2-5x+4\right)-3\left(3x^2-5x+4\right)\right]:\left(2x-3\right)\)

\(=\left[\left(3x^2-5x+4\right)\left(2x-3\right)\right]:\left(2x-3\right)\)

\(=3x^2-5x+4\)

\(e,\)

\(\left(6x^3-5x^2+4x-1\right):\left(2x^2-x+1\right)\)

\(=\left(6x^3-3x^2-2x^2+3x+x-1\right):\left(2x^2-x+1\right)\)

\(=\left[\left(6x^3-3x^2+3x\right)-\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\)

\(=\left[3x\left(2x^2-x+1\right)-\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\)

\(=\left[\left(2x^2-x+1\right)\left(3x-1\right)\right]:\left(2x^2-x+1\right)\)

\(=3x-1\)

\(f,\)

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left(x^3-4x^2+2x^2+3x-8x+6\right):\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)-\left(4x^2+8x\right)+\left(3x+6\right)\right]\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

\(=x^2-4x+3\)

\(g,\)

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left(x^3+2x^2-4x^2-8x+3x+6\right):\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)-\left(4x^2+8x\right)+\left(3x+6\right)\right]:\left(x+2\right)\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]:\left(x+2\right)\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

\(=x^2-4x+3\)

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