a) \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right):\dfrac{x}{x+2019}\)
\(A=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x+2019}{x}\)
\(A=\left(\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x+2019}{x}\)
\(A=\left(\dfrac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x+2019}{x}\)
\(A=\left(\dfrac{\left(x^2+2x+1\right)-\left(x^2-2x+1\right)+x^2-4x-1}{x^2-1}\right)\cdot\dfrac{x+2019}{x}\)
\(A=\left(\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{x^2-1}\right)\cdot\dfrac{x+2019}{x}\)
\(A=\left(\dfrac{x^2-1}{x^2-1}\right)\cdot\dfrac{x+2019}{x}\)
\(A=1\cdot\dfrac{x+2019}{x}\)
\(A=\dfrac{x+2019}{x}\)
b) \(A=2023\)
\(\Rightarrow\dfrac{x+2019}{x}=2023\)
\(\Leftrightarrow\dfrac{x+2019}{x}=\dfrac{2023x}{x}\)
\(\Leftrightarrow x+2019=2023x\)
\(\Leftrightarrow2019=2023x-x\)
\(\Leftrightarrow2019=2022x\)
\(\Leftrightarrow x=\dfrac{2019}{2022}=\dfrac{673}{674}\)
\(a,\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}+\dfrac{x^2-4x-1}{x^2-1}\right):\dfrac{x}{x+2019}\\ =\left(\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}+\dfrac{x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{x}{x+2019}\\ =\dfrac{x^2+2x+1-\left(x^2-2x+1\right)+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x}{x+2019}\\ =\dfrac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}:\dfrac{x}{x+2019}\\ =\dfrac{x^2-1}{\left(x-1\right)\left(x+1\right)}.\dfrac{x+2019}{x}\\ =1.\dfrac{x+2019}{x}\\ =\dfrac{x+2019}{x}\)
`b,` Để `A=2023` thì
\(\dfrac{x+2019}{x}=2023\\ \Leftrightarrow x+2019=2023x\\ \Leftrightarrow x-2023x=-2019\\ \Leftrightarrow-2022x=-2019\\ \Leftrightarrow x=\dfrac{673}{674}\)
Vậy để `A=2023` thì `x=673/674`
