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a) Đa thức thương x 2 – 6x + 9.
b) Đa thức thương 2 x 2 – 5.
c) Đa thức thương x 2 + 4x + 3 và đa thức dư -12.
d) Đa thức x + 5 và đa thức dư x – 4.
\(a,=10x^3-5x^2+5x\\ b,=x^3+27\\ c,=\dfrac{5}{2}xy-1-\dfrac{1}{2}y\\ d,=\left(2x^3-10x^2-11x^2+55x+12x-60\right):\left(x-5\right)\\ =\left[2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\right]:\left(x-5\right)\\ =2x^2-11x+12\)
a) \(\dfrac{3}{4xy}+\dfrac{5x}{2x^2z}+\dfrac{7}{6yz^2}\) (MSC: \(12x^2yz^2\))
\(=\dfrac{3\cdot3xz^2}{4xy\cdot3xz^2}+\dfrac{5x\cdot6yz}{2x^2z\cdot6yz}+\dfrac{7\cdot2x^2}{6yz^2\cdot2x^2}\)
\(=\dfrac{9xz^2}{12x^2yz^2}+\dfrac{30xyz}{12x^2yz^2}+\dfrac{14x^2}{12x^2yz^2}\)
\(=\dfrac{9xz^2+30xyz+14x^2}{12x^2yz^2}\)
\(=\dfrac{x\left(9z^2+30yz+14x\right)}{12x^2yz^2}\)
\(=\dfrac{9z^2+30yz+14x}{12x^2yz^2}\)
b) \(\dfrac{x^2}{x^2+3x}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x^2}{x\left(x+3\right)}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}\)
\(=\dfrac{x+3}{x+3}+\dfrac{3}{x}\)
\(=1+\dfrac{3}{x}\)
\(=\dfrac{x}{x}+\dfrac{3}{x}\)
\(=\dfrac{x+3}{x}\)
a: \(=\dfrac{3\cdot3\cdot xz^2+5x\cdot6\cdot y+7\cdot x^2\cdot2}{12x^2yz^2}=\dfrac{9xz^2+30xy+14x^2}{12x^2yz^2}\)
\(=\dfrac{9z^2+30y+14x}{12xyz^2}\)
b: \(=\dfrac{x}{x+3}+\dfrac{3}{x+3}+\dfrac{3}{x}=1+\dfrac{3}{x}=\dfrac{x+3}{x}\)
a) \(\frac{2x^3+x^2+x+6}{x^2-x+2}=\frac{\left(2x+3\right)\left(x^2-x+2\right)}{x^2-x+2}=2x+3\)
b) \(\frac{x}{x-3}-\frac{5x^2+27}{x^2-9}+\frac{x-9}{x+3}\)
\(=\frac{x}{x-3}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x-9}{x+3}\)
\(=\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-9\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}-\frac{5x^2+27}{\left(x-3\right)\left(x+3\right)}+\frac{x^2-12x+27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x-\left(5x^2+27\right)+x^3-12x+27}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x^2-9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{-3x}{x-3}\)