Question

Answer 1

Bài `3`

\(a,B=\dfrac{2}{x^2-5x+6}:\dfrac{x+2}{x-2}\\ =\dfrac{2}{x^2-2x-3x+6}\cdot\dfrac{x-2}{x+2}\\ =\dfrac{2}{x\left(x-2\right)-3\left(x-2\right)}\cdot\dfrac{x-2}{x+2}\\ =\dfrac{2}{\left(x-2\right)\left(x-3\right)}\cdot\dfrac{x-2}{x+2}\\ =\dfrac{2}{\left(x-3\right)\left(x+2\right)}\)

`b,` Khi `x=5` ta có :

\(\dfrac{2}{\left(x-3\right)\left(x+2\right)}\Rightarrow\dfrac{2}{\left(5-3\right)\left(5+2\right)}=\dfrac{2}{2\cdot7}=\dfrac{2}{14}=\dfrac{1}{7}\)

Answer 2

3:

a: \(B=\dfrac{2}{x^2-5x+6}:\dfrac{x+2}{x-2}\)

\(=\dfrac{2}{\left(x-2\right)\left(x-3\right)}\cdot\dfrac{x-2}{x+2}\)

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

b: khi x=5 thì \(B=\dfrac{2}{\left(5+2\right)\left(5-3\right)}=\dfrac{1}{7}\)