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Ta có: \(\dfrac{3x^4-8x^3-10x^2+6x-3}{3x^2-2x+1}\)
\(=\dfrac{3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5+2x+2}{3x^2-2x+1}\)
\(=x^2-2x-5+\dfrac{2x+2}{3x^2-2x+1}\)
a: \(=\dfrac{x^4-6x^3+12x^2-14x+3}{x^2-4x+1}\)
\(=\dfrac{x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3}{x^2-4x+1}\)
\(=x^2-2x+3\)
b: \(=\dfrac{x^5-3x^4+5x^3-x^2+3x-5}{x^2-3x+5}=x^2-1\)
c: \(=\dfrac{2x^4-5x^3+2x^2+2x-1}{x^2-x-1}\)
\(=\dfrac{2x^4-2x^3-2x^2-3x^3+3x^2+3x+x^2-x-1}{x^2-x-1}\)
\(=2x^2-3x+1\)
1.
a) \(\left(-2x^3\right)\)\(\left(x^2+5x-\frac{1}{2}\right)\) = \(-2x^5\)\(-10x^4\) \(+x^3\)
b) (\(6x^3-7x^2\)\(-x+2\))\(:\left(2x+1\right)\)=\(3x^2-5x+2\)
2.
a) 9x(3x-y) + 3y (y-3x)=9x(3x-y)-3y(3x-y)
= (9x-3y)(3x-y)
= 3(3x-y)(3x-y)
= 3(3x-y)^2
b) \(x^3-3x^2\)\(-9x+27\)= \(\left(x^3-3x^2\right)\)\(-\left(9x-27\right)\)
= \(x^2\left(x-3\right)\)\(-9\left(x-3\right)\)
= \(\left(x^2-9\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)\left(x-3\right)\)
= \(\left(x+3\right)\left(x-3\right)^2\)
Bài 1 ) a ) \(\left(-2x^3\right)\left(x^2+5x-\frac{1}{2}\right)\)
\(=-2x^5-10x^4+x^3\)
b ) \(\left(6x^3-7x^2+x+2\right):\left(2x+1\right)\)
\(=3x^2-5x+2\)
2 ) a ) \(9x\left(3x-y\right)+3y\left(y-3x\right)\)
\(=9x\left(3x-y\right)-3y\left(3x-y\right)\)
\(=\left(3x-y\right)\left(9x-3y\right)\)
\(=3\left(3x-y\right)\left(x-y\right)\)
b ) \(x^3-3x^2-9x+27\)
\(=\left(x^3-3x^2\right)-\left(9x-27\right)\)
\(=x^2\left(x-3\right)-9\left(x-3\right)\)
\(=\left(x^2-9\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x+3\right)\left(x-3\right)\)
