Question

1.Dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng

d)\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)

Answer 1

pt <=> \(\left(x^2-x-2\right)\left(x-1\right)=\left(x+1\right)\left(x^2-3x+2\right)\)

\(\Leftrightarrow\left(x^2+x-2x-2\right)\left(x-1\right)=\left(x+1\right)\left(x^2-x-2x+2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x-1\right)=\left(x+1\right)\left(x-2\right)\left(x-1\right)\)(Đúng \(\forall x\) )

Answer 2

Ta có:

\(\left(x^2-x-2\right)\left(x-1\right)\)

= \(\left(x^2-2x+x-2\right)\left(x-1\right)\)

= \([\left(x^2-2x)+(x-2\right)]\left(x-1\right)\)

= \([x\left(x-2)+(x-2\right)]\left(x-1\right)\)

= \(\left(x-2\right)\left(x+1\right)\left(x-1\right)\) (1)

Lại có:

\((x^2-3x+2)\left(x+1\right)\)

= \((x^2-2x-x+2)\left(x+1\right)\)

= \([(x^2-2x)-(x-2)]\left(x+1\right)\)

= \([x(x-2)-(x-2)]\left(x+1\right)\)

= \(\left(x-2\right)\left(x-1\right)\left(x+1\right)\) (2)

Từ (1), (2)

=> \(\left(x^2-x-2\right)\left(x-1\right)\) = \((x^2-3x+2)\left(x+1\right)\)

=> \(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)

Answer 3

Ta có :

\(\dfrac{x^2-x-2}{x+1}=\dfrac{x^2-3x+2}{x-1}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)}{x+1}=\dfrac{\left(x-1\right)\left(x-2\right)}{x-1}\)

\(\Leftrightarrow x-2=x-2\)

\(\Rightarrow\) 2 phân thức bằng nhau (đpcm)

Answer 4

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

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

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