ĐKXĐ x khác 3,-1/3

\(A=\frac{3x^3-9x^2-5x^2+15x-12x+36}{3x^3-9x^2-10x^2+30x+3x-9}\)

   \(=\frac{3x^2\left(x-3\right)-5x\left(x-3\right)-12\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

 \(=\frac{\left(x-3\right)\left(3x^2-5x-12\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

\(=\frac{3x^2-5x-12}{3x^2-10x+3}=\frac{\left(x-3\right)\left(3x+4\right)}{\left(x-3\right)\left(3x-1\right)}\)

\(=\frac{3x+4}{3x-1}\)

b,với ĐKXĐ ta có \(A=0\Leftrightarrow\frac{3x+4}{3x-1}=0\Leftrightarrow3x+4=0\Leftrightarrow x=\frac{-4}{3}\left(tm\right)\)

c,\(\frac{3x+4}{3x-1}=\frac{3x-1+5}{3x-1}=1+\frac{5}{3x-1}\)

để A thuộc z thì \(\frac{5}{3x-1}\in Z\Rightarrow3x-1\inƯ\left(5\right)\)                   đến đây bạn tìm ước của 5 rồi tự giải nhé

Cho mình hỏi dòng dấu = thứ 4 làm sao vậy. ko hiểu

Cảm ơn câu trả lời của bạn Nguyễn Thị Huyền Anh, mình hiểu rồi

sai đề rồi bạn ơi

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)

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

Còn bài b bạn tự làm nhé

Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)

\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)

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

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

Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

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

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

\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)

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

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

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:

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

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:

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