Tử: \(2x^3-7x^2-12x+45\)
\(=2x^3-12x^2+5x^2+18x-30+45\)
\(=2x^3-12x^2+18x+5x^2-30x+45\)
\(=2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)\)
\(=\left(2x+5\right)\left(x-3\right)^2\) \(\left(1\right)\)
Mẫu: \(3x^3-19x^2+33x-9\)
\(=3x^3-18x^2-x^2+27x+6x-9\)
\(=3x^3-18x^2+27x-x^2+6x-9\)
\(=3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)\)
\(=\left(3x-1\right)\left(x-3\right)^2\) \(\left(2\right)\)
Từ (1) và (2) ta được: \(\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)
(Nghĩ vậy chứ cũng không chắc lắm)
Đặt \(A=\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
\(=\frac{2x^2-6x+5x-15}{3x^2-x-9x+3}\)
\(=\frac{2x\left(x-3\right)+5\left(x-3\right)}{x\left(3x-1\right)-3\left(3x-1\right)}\)
\(=\frac{\left(x-3\right)\left(2x+5\right)}{\left(3x-1\right)\left(x-3\right)}=\frac{2x+5}{3x-1}\)
@Băng Băng 2k6 Đúng rồi đấy ! Học giỏi lắm !
