Question

Answer 1

`a) [ x + 1 ] / [ x + 2 ] + 5 / [ x - 2 ] = 4 / [ x^2 - 4 ] + 1`      `ĐK: x \ne +-2`

`<=> [ ( x + 1 ) ( x - 2 ) + 5 ( x + 2 ) ] / [ ( x - 2 ) ( x + 2 ) ] = [ 4 + x^2 - 4 ] / [ ( x - 2 ) ( x + 2 ) ]`

  `=> x^2 - 2x + x - 2 + 5x + 10 = 4 + x^2 - 4`

`<=> x^2 - x^2 - 2x + x + 5x = 4 - 4 + 2 - 10`

`<=> 4x = -8`

`<=> x = -2`  (t/m)

Vậy `S = \emptyset`

Answer 2

\(a,\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)+5\left(x+2\right)-4-x^2+4}{x^2-4}=0\) \(\left(dk:x\ne\pm2\right)\)

\(\Leftrightarrow x^2-2x+x-2+5x+10-4-x^2+4=0\)

\(\Leftrightarrow4x+8=0\)

\(\Leftrightarrow x=-2\left(l\right)\)

Vậy \(S=\varnothing\)

Answer 3

\(ĐK:x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x-2\right)+5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4+\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow\left(x+1\right)\left(x-2\right)+5\left(x+2\right)=4+\left(x^2-4\right)\)

\(\Leftrightarrow x^2-2x+x-2+5x+10=4+x^2-4\)

\(\Leftrightarrow3x=-8\)

\(\Leftrightarrow x=-\dfrac{8}{3}\left(tm\right)\)

Vậy \(S=\left\{-\dfrac{8}{3}\right\}\)