Question

Answer 1

a: \(Q=\dfrac{x-1}{x+2}+\dfrac{5x-2}{x^2-4}\)

\(=\dfrac{x-1}{x+2}+\dfrac{5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-2\right)+5x-2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2-3x+2+5x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}\)

b: Sửa đề: Tính giá trị của Q khi x=-3

Thay x=-3 vào Q, ta được:

\(Q=\dfrac{-3}{-3-2}=\dfrac{-3}{-5}=\dfrac{3}{5}\)

c: Q=1

=>\(\dfrac{x}{x-2}=1\)

=>x=x-2

=>0=-2(vô lý)