https://olm.vn/hoi-dap/tim-kiem?q=%7C2x-9%7C+%7C3x-13%7C-%7C14-5x%7C+%7C19-(4x+12)%7C=56t%C3%ACm+x&id=191953

\(\Leftrightarrow\left[{}\begin{matrix}14-19x=7x-15\\19x-14=7x-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}26x=29\\12x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{26}\\x=-\dfrac{1}{12}\end{matrix}\right.\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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)

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)

Ta có tử bằng: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)²(2x+5)                   (1)

Ta có mẫu bằng:3x³-19x²+33x-9

                        =(3x³-x²)-(19x²-6x)+(27x-9)

                        =x²(3x-1)-6x(3x-1)+9(3x-1)

                        =(3x-1)(x²-6x+9)

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

Thay (1) và (2) vào phân thức ,ta có:

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}=\frac{2x+5}{3x-1}\)

Ta có: \(C=\frac{3x^2-7x^2-12+45}{3x^3-19x^2+33x-9}\)    ĐKXĐ: x khác 3, 1/3 

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

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

Để C>0, ta có:

-5/2<x<1/3 (thỏa mãn ĐKXĐ) 

Bạn xem lại cái đề bài đi :))))) 

Lời giải:

Ta có:

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

Xét \(\text{TS}=2x^2(x-3)-x(x-3)-15(x-3)\)

\(=(x-3)(2x^2-x-15)=(x-3)[2x(x-3)+5(x-3)]\)

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

Xét \(\text{MS}=3x^2(x-3)-10x(x-3)+3(x-3)\)

\(=(x-3)(3x^2-10x+3)=(x-3)[3x(x-3)-(x-3)]\)

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

Do đó:

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(x-3)^2(3x-1)}=\frac{2x+5}{3x-1}\)

Xét tử thức ta có

2x³-7x²-12x+45

= 2x³+5x²-12x²-30x+18x+45

= x²(2x+5)-6x(2x+5)+9(2x+5)

= (2x+5)(x²-6x+9)

= (2x+5)(x-3)² (1)

Xét mẫu thức ta có

3x³-19x²+33x-9

= 3x³-x²-18x²+6x+27x-9

= x²(3x-1)-6x(3x-1)+9(3x-1)

= (3x-1)(x²-6x+9)

= (3x-1)(x-3)² (2)

Thay (1) và (2) vào A ta được\(A=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)

a, Để phân thức trên có nghĩa thì:

      \(3x^3-19x^2+33x-9\ne0\)

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

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

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

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

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

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

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

1. \(x^2+2x-15=0\)

\(\Rightarrow x^2+2x+1^2-16=0\)

\(\Rightarrow\left(x+1\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\).

2. \(x^2-7x-44=0\)

\(\Rightarrow x^2-2.x.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{49}{4}-44=0\)

\(\Rightarrow\left(x-\dfrac{7}{4}\right)^2=\left(\dfrac{15}{2}\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{7}{4}=\dfrac{15}{2}\\x-\dfrac{7}{4}=-\dfrac{15}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{37}{4}\\x=\dfrac{-23}{4}\end{matrix}\right.\).

3.4 Tương tự.

2) hãy dành 5(s)

\(x^2-7x-44=0\Rightarrow\left(x^2+4x\right)-\left(11x+44\right)=0\)

\(x\left(x+4\right)-11\left(x+4\right)=0\)

\(\left(x+4\right)\left(x-11\right)=0\)\(\left[{}\begin{matrix}x=-4\\x=11\end{matrix}\right.\)