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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

a, Để phân thức trên có nghĩa thì:

      \(3x^3-19x^2+33x-9\ne0\)

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

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

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

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

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

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

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

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:

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

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

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 @_@

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