Question

Cho P=\(\left(\dfrac{a^2+3a+2}{a^2+a-2}-\dfrac{a^2+4}{a^2-1}\right):\left(\dfrac{1}{a+1}+\dfrac{1}{a-1}\right)\)

a)Tìm điều kiện xác định và rút gọn P

b)Tìm a để |P|=\(\dfrac{2}{3}\)

c)Tìm a để \(\dfrac{1}{P}-\dfrac{a+1}{8}=1\)

Answer 1

a: ĐKXĐ: a<>-1; a<>1; a<>-2;a<>0

\(P=\left(\dfrac{\left(a+2\right)\left(a+1\right)}{\left(a+2\right)\left(a-1\right)}-\dfrac{a^2+4}{\left(a-1\right)\left(a+1\right)}\right):\dfrac{a-1+a+1}{\left(a+1\right)\left(a-1\right)}\)

\(=\left(\dfrac{a+1}{a-1}-\dfrac{a^2+4}{\left(a-1\right)\left(a+1\right)}\right)\cdot\dfrac{\left(a+1\right)\left(a-1\right)}{2a}\)

\(=\dfrac{a^2+2a+1-a^2-4}{2a}=\dfrac{2a-3}{2a}\)

b: |P|=2/3 thì P=2/3 hoặc P=-2/3

P=2/3 thì 2a-3/2a=2/3

=>6a-9=4a

=>2a=9

=>a=9/2

P=-2/3 thì 2a-3/2a=-2/3

=>6a-9=-4a

=>10a=9

=>a=9/10