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\(\frac{2\left(x-2\right)\left(x-3\right)}{\left(3x^2-27\right)\left(x-2\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{\left(x-2\right)3\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ne1;x\ne-\dfrac{3}{2}\)
Ta có: \(\dfrac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\dfrac{\left(x-1\right)^2\left(3x-1\right)}{\left(x-1\right)^2\left(2x+3\right)}=\dfrac{3x-1}{2x+3}\)
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a: \(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)
b: \(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)
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Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+2x^2-x-2\)
\(=x^3+3x^2+2x-1x^2-3x-2\)
\(=x\left(x^2+3x+2\right)-1\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x^2+3x+2\right)\)
\(=\left(x-1\right)\left(x^2+x+2x+2\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(x^3+3x+2\)
\(=x^3+2x^2-2x^2-4x+x+2\)
\(=\left(x+2\right)x^2-2\left(x^2+2x\right)+x+2\)
\(=\left(x+2\right)x^2-2\left(x^2+2x\right)1\left(x+2\right)\)
\(=\left(x^2-2x+1\right)\left(x+2\right)\)
\(=\left(x-1\right)^2\left(x+2\right)\)