a, Để phân thức trên có nghĩa thì:

      \(3x^3-19x^2+33x-9\ne0\)

 \(\Rightarrow3x^3-9x^2-10x^2+30x+3x-9\ne0\)

\(\Rightarrow3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)\ne0\)

\(\Rightarrow\left(x-3\right)\left(3x^2-10x+3\right)\ne0\)

\(\Rightarrow\left(x-3\right).\left[3x^2-9x-x+3\right]\ne0\)

\(\Rightarrow\left(x-3\right)\left[3x\left(x-3\right)-\left(x-3\right)\right]\ne0\)

\(\Rightarrow\left(x-3\right)^2.\left(3x-1\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x-3\ne0\\3x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne3\\x\ne\frac{1}{3}\end{cases}}}\)

a, Ra đáp án luôn nha

B=(2x+5)/(3x-1)

b,Để B>0 thì 2x+5 và 3x-1 phải cùng dấu 

Đáp án : x khác 0;-1;-2

a, Ra đáp án luôn nha
B=(2x+5)/(3x-1)
b,Để B>0 thì 2x+5 và 3x-1 phải cùng dấu
Đáp án : x khác 0;-1;-2

chúc bn hok tốt @_@

a, mk làm đáp án luôn đó

B=(2x+5)/(3x-1)

b,Để B>0 thì 2x+5 và 3x-1 phải cùng dấu 

=> : x khác 0;-1;-2

\(B=\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(=\dfrac{2x^3+5x^2-12x^2-30x+18x+45}{3x^3-x^2-18x^2+6x+27x-9}\)

\(=\dfrac{\left(2x^3+5x^2\right)-\left(12x^2+30x\right)+\left(18x+45\right)}{\left(3x^3-x^2\right)-\left(18x^2-6x\right)+\left(27x-9\right)}\)

\(=\dfrac{x^2\left(2x+5\right)-6x\left(2x+5\right)+9\left(2x+5\right)}{x^2\left(3x-1\right)-6x\left(3x-1\right)+9\left(3x-1\right)}\)

\(=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)

\(=\dfrac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}\)

ĐKXĐ : \(\left\{{}\begin{matrix}3x-1\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{1}{3}\\x\ne3\end{matrix}\right.\)

\(a,B=\dfrac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\dfrac{2x+5}{3x-1}\)

b,Để \(B>0\)

\(\Leftrightarrow\dfrac{2x+5}{3x-1}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+5>0\\3x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+5< 0\\3x-1< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-\dfrac{5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< -\dfrac{5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{5}{2}\end{matrix}\right.\) thì B > 0

a) ĐKXĐ:\(x\ne\dfrac{1}{3};x\ne3\)

\(B=\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(B=\dfrac{\left(2x^3-12x^2+18x\right)+\left(5x^2-30x+45\right)}{\left(3x^3-18x^2+27x\right)-\left(x^2-6x+9\right)}\)

\(B=\dfrac{2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)}{3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)}\)

\(B=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)

\(B=\dfrac{2x+5}{3x-1}\)

b) Để \(B>0\Leftrightarrow\dfrac{2x+5}{3x-1}>0\Leftrightarrow2x+5\)và \(3x-1\) cùng dấu

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+5>0\\3x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+5< 0\\3x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{-5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{-5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{5}{2}\end{matrix}\right.\)

Câu a trả lời chi tiết giúp mình ạ

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(=\frac{\left(x-\frac{2}{5}\right)\left(x+3\right)}{\left(x+\frac{1}{3}\right)\left(x+3\right)}\)

\(=\frac{x-\frac{2}{5}}{x+\frac{1}{3}}\)

=\(\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

=\(\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

=\(\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

=\(\frac{2x^2-6x+5x-15}{3x^2-9x-x+3}\)

=\(\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)

=\(\frac{2x+5}{3x-1}\)

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)

\(=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

\(=\frac{\left(2x^2-x-15\right)\left(x-3\right)}{\left(3x^2-10x+3\right)\left(x-3\right)}\)

\(=\frac{2x^2-x-15}{3x^2-10x+3}\)

2x³ - 7x² - 12x + 45 = 2x³ - 6x² - x² + 3x - 15x + 45

                          =  2x²(x - 3) - x(x - 3) - 15(x - 3)

                          = (x - 3)(2x² - x - 15)

                          = (x - 3)(2x² - 6x + 5x - 15)

                          = (x -  3)((2x(x - 3) + 5(x - 3))

                         =  (x - 3)²(2x + 5)

3x³ - 19x² +33x - 9 = 3x³ -9x² -10x² + 30x +3x - 9

                              = 3x²(x - 3) - 10x(x - 3) + 3(x - 3)

                             = (x - 3)(3x² - 10x + 3)

                            = (x - 3)(3x² -9x - x +3)

                            = (x - 3)((3x(x-3) - (x - 3))

                           =(x - 3)²(3x - 1)