Question

Giải pt (a,b là các tham số)

a, a(ax+1)=x(a+2)+2

b, \(\dfrac{x-a}{a+1}+\dfrac{x-1}{a-1}=\dfrac{2a}{1-a^2}\)

Answer 1

\(\dfrac{x-a}{a+1}+\dfrac{x-1}{a-1}=\dfrac{2a}{1-a^2}\) (ĐK: \(a\ne\pm1\))

\(\Rightarrow\dfrac{\left(x-a\right)\left(a-1\right)}{a^2-1}+\dfrac{\left(x-1\right)\left(a+1\right)}{a^2-1}+\dfrac{2a}{a^2-1}=0\)

\(\Rightarrow\dfrac{ax-x-a^2+a+ax+x-a-1+2a}{a^2-1=0}\)

\(\Rightarrow\dfrac{2ax-a^2+2a-1}{a^2-1}=0\)

\(\Rightarrow2ax-\left(a^2-2a+1\right)=0\)

\(\Rightarrow2ax-\left(a-1\right)^2=0\)

Với a =0 , ta có đẳng thưc sai

Với \(a\ne0\), ta được :

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