Question

Cho \(A=\left(\dfrac{a+2}{a+1}-\dfrac{a-2}{a-1}\right).\dfrac{a+1}{a};B=\dfrac{3}{a^2-1}\)

Tìm a để A=2B

Answer 1

ĐKXĐ: \(a\notin\left\{1;-1;0\right\}\)

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

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

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

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

A=2B

=>\(\dfrac{2}{a-1}=\dfrac{2\cdot3}{a^2-1}\)

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

=>\(\dfrac{1}{a-1}=\dfrac{3}{\left(a-1\right)\left(a+1\right)}\)

=>\(\dfrac{a+1}{\left(a-1\right)\left(a+1\right)}=\dfrac{3}{\left(a-1\right)\left(a+1\right)}\)

=>a+1=3

=>a=2(nhận)