Thực hiện phép chia
( -3x^3 + 5x^2 - 9 + 15 ) : ( -3x + 5 )
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\(=\dfrac{5x^6-3x^3+x^2}{9x^4}=\dfrac{5}{9}x^2-\dfrac{1}{3x}+\dfrac{1}{9x^2}\)
a: \(=\dfrac{5\left(x+2\right)}{10xy^2}\cdot\dfrac{12x}{x+2}=\dfrac{60x}{10xy^2}=\dfrac{6}{y^2}\)
b: \(=\dfrac{x-4}{3x-1}\cdot\dfrac{3\left(3x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{3}{x+4}\)
c: \(=\dfrac{2\left(2x+1\right)}{\left(x+4\right)^2}\cdot\dfrac{\left(x+4\right)}{3\left(x+3\right)}=\dfrac{2\left(2x+1\right)}{3\left(x+3\right)\left(x+4\right)}\)
d: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
\(\frac{{9{x^2} + 5x + x}}{{3x}} = \frac{{9{x^2} + 6x}}{{3x}} = \frac{{9{x^2}}}{{3x}} + \frac{{6x}}{{3x}} = 3x + 2\)
\(\frac{{2{x^2} - 3x - 2}}{{2 - x}} = \frac{{2{x^2} - 3x - 2}}{{ - x + 2}} = - 2x - 1\)
Câu 1 :
\(2x^2\left(3x-5x^3\right)+10x^5-5x^3\)
\(=6x^3-10x^5+10x^5-5x^3\)
\(=x^3\)
Câu 2 :
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x-9\right)\left(x+3\right)\)
\(=x^3+3^3+\left(x^2+3x-9x-27\right)\)
\(=x^3+27+x^2-6x-27\)
\(=x^3+x^2-6x\)
\(\left(-3x^3+5x^2-9+15\right):\left(-3x+5\right)\)
\(=[x^2.\left(-3x+5\right)+3.\left(-3+5\right)]:\left(-3x+5\right)\)
\(=\left(x^2+3\right).\left(-3+5\right):\left(-3+5\right)\)
\(=x^2+3\)