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\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\frac{\left(x-\frac{2}{5}\right)\left(x+3\right)}{\left(x+\frac{1}{3}\right)\left(x+3\right)}\)
\(=\frac{x-\frac{2}{5}}{x+\frac{1}{3}}\)
=\(\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)
=\(\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)
=\(\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)
=\(\frac{2x^2-6x+5x-15}{3x^2-9x-x+3}\)
=\(\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)
=\(\frac{2x+5}{3x-1}\)
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\(A=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)
\(B=\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
\(=\dfrac{2x^3+5x^2-12x^2-30x+18x+45}{3x^3-x^2-18x^2+6x+27x-9}\)
\(=\dfrac{\left(2x^3+5x^2\right)-\left(12x^2+30x\right)+\left(18x+45\right)}{\left(3x^3-x^2\right)-\left(18x^2-6x\right)+\left(27x-9\right)}\)
\(=\dfrac{x^2\left(2x+5\right)-6x\left(2x+5\right)+9\left(2x+5\right)}{x^2\left(3x-1\right)-6x\left(3x-1\right)+9\left(3x-1\right)}\)
\(=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)
\(=\dfrac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}\)
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
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x+5>0\\3x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x+5< 0\\3x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{-5}{2}\\x>\dfrac{1}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{-5}{2}\\x< \dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -\dfrac{5}{2}\end{matrix}\right.\)
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(2x+5\right)\left(x-3\right)^2}{\left(3x-1\right)\left(x-3\right)^2}=\frac{2x+5}{3x-1}\)
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:
\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{\left(x-3\right)^2\left(2x+5\right)}{\left(x-3\right)^2\left(3x-1\right)}=\frac{2x+5}{3x-1}\)
\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)
\(=\frac{2x+5}{3x-1}\)
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, 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