Question

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

a,tìm điều kiện xácđịnh

b,rút gọn PT

tính giá trị của P khi a=2

Answer 1

a) \(ĐKXĐ:a\ne\pm1\)

b) \(P=\left(\dfrac{a+1}{2a-2}+\dfrac{1}{2-2a^2}\right)\cdot\dfrac{2a+2}{a+2}\)

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

\(=\left(\dfrac{a+1}{2\left(a-1\right)}-\dfrac{1}{2\left(a-1\right)\left(a+1\right)}\right)\cdot\dfrac{2\left(a+1\right)}{a+2}\)

\(=\dfrac{\left(a+1\right)\left(a-1\right)-1}{2\left(a-1\right)\left(a+1\right)}\cdot\dfrac{2\left(a+1\right)}{a+2}\)

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

\(=\dfrac{a^2-2}{a^2+a-2}\)

Khi a = 2 thì :

\(P=\dfrac{2^2-2}{2^2+2-2}=\dfrac{2}{4}=\dfrac{1}{2}\)

p/s: check lại hộ tui nhá =)))