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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: \(=\dfrac{5\left(x-1\right)}{3\left(x+1\right)}\cdot\dfrac{x+1}{x-1}=\dfrac{5}{3}\)
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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 :
\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x-9\right)\left(x+3\right)\)
\(=x^3+3^3+\left(x^2+3x-9x-27\right)\)
\(=x^3+27+x^2-6x-27\)
\(=x^3+x^2-6x\)
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1:=x^3-27-x^2-3=x^3-x^2-30
2: =x-2+125x^3+150x^2+60x+8
=125x^3+150x^2+61x+6
3: \(=2xy-5y+5y=2xy\)
4: =25x-10x^2+15x
=-10x^2+40x
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a) \(\dfrac{x^2+2x}{x+2}=\dfrac{x\left(x+2\right)}{x+2}=x\)
b) \(\dfrac{5x+4-3\left(x-2\right)}{3\left(x+5\right)}=\dfrac{5x+4-3x+6}{3\left(x+5\right)}=\dfrac{2x+10}{3\left(x+5\right)}=\dfrac{2\left(x+5\right)}{3\left(x+5\right)}=\dfrac{2}{3}\)
\(\left(-3x^3+5x^2-9+15\right):\left(-3x+5\right)\)
\(=[x^2.\left(-3x+5\right)+3.\left(-3+5\right)]:\left(-3x+5\right)\)
\(=\left(x^2+3\right).\left(-3+5\right):\left(-3+5\right)\)
\(=x^2+3\)