Question

[trần anh tú]

1, tính

a,\(\dfrac{x+1}{x+2}:\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\)

b,\(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)

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

d,\(\dfrac{bc}{\left(a-b\right).\left(a-c\right)}+\dfrac{ac}{\left(b-a\right).\left(b-c\right)}+\dfrac{ab}{\left(c-a\right).\left(a-b\right)}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(=\dfrac{x+1}{x+2}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+1}{x+3}=\dfrac{\left(x+1\right)^2}{\left(x+2\right)^2}\)

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

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

c: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)

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

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