\(\left\{{}\begin{matrix}\dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\\dfrac{x-3}{x}\\\dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\end{matrix}\right.\)

Vậy \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)

\(ĐK:x\ne0;x\ne\pm1\\ \dfrac{x^2-2x-3}{x^2+x}=\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}\\ \dfrac{x^2-4x+3}{x^2-x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}=\dfrac{x-3}{x}\)

Do đó 3 phân thức trên bằng nhau

\(\text{Ta có : }\dfrac{x^2-2x-3}{x^2+x}\\ =\dfrac{x^2+x-3x-3}{x\left(x+1\right)}\\ =\dfrac{\left(x^2+x\right)-\left(3x+3\right)}{x\left(x+1\right)}\\ \\ =\dfrac{x\left(x+1\right)-3\left(x+1\right)}{x\left(x+1\right)}\\ \\ =\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\text{ }\left(1\right)\)

\(\dfrac{x^2-4x+3}{x^2-x}\\ =\dfrac{x^2-x-3x+3}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x^2-x\right)-\left(3x-3\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{\left(x-3\right)\left(x-1\right)}{x\left(x-1\right)}\\ \\ =\dfrac{x-3}{x}\text{ }\text{ }\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)

Vậy 3 phân thức \(\dfrac{x^2-2x-3}{x^2+x};\dfrac{x-3}{x};\dfrac{x^2-4x+3}{x^2-x}\) bằng nhau

Giả sử :

\(\dfrac{x^2-2x-3}{x^2+x}=\dfrac{x-3}{x}=\dfrac{x^2-4x+3}{x^2-x}\)

\(\Leftrightarrow\) \(\dfrac{\left(x+1\right)\left(x-3\right)}{x\left(x+1\right)}=\dfrac{x-3}{x}=\dfrac{\left(x-1\right)\left(x-3\right)}{x\left(x-1\right)}\)

\(\Leftrightarrow\dfrac{x-3}{x}=\dfrac{x-3}{x}=\dfrac{x-3}{x}\)

Vậy 3 thức trên bằng nhau

a) \(\dfrac{1}{x-a};\dfrac{2}{x-b}\)

Theo đề bài ta có :

\(\left(x-a\right)\left(x-b\right)=x^2-5x+6\)

\(\Leftrightarrow\left(x-a\right)\left(x-b\right)=\left(x-2\right)\left(x-3\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)

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

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

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

\(\Rightarrow\frac{x^2+5}{x^3-3x-2}=\frac{ax^2+\left(2a+b\right)x+\left(a-2b\right)}{x^3-3x-2}\)

Đồng nhất hệ số, ta có :

\(\hept{\begin{cases}a=1\\2a+b=0\\a-2b=5\end{cases}\Rightarrow\hept{\begin{cases}a=1\\b=-2\end{cases}}}\)

cái thứ 2 tương tự

=.= tương tự là seo????

giup e vs ah