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điều kiện \(x\ne2;x\ne-2\)
\(\frac{\left(x+2\right)^2.P}{x^2-4}=\frac{\left(x-1\right).Q}{x^2-4}\)
\(\left(x+2\right)^2.P=\left(x-1\right)Q\)
\(P=x-1\)
\(Q=\left(x+2\right)^2\)
x + 2 P x - 2 = x - 1 Q x 2 - 4
⇒ x + 2 . P . x 2 - 4 = x - 2 x - 1 . Q
Hay (x + 2)(x – 2)(x + 2).P = (x – 2)(x – 1).Q
Chọn P = (x – 1) thì Q = x + 2 2
a) \(\dfrac{\left(x+2\right)P}{x-2}=\dfrac{\left(x-1\right)Q}{x^2-4}\)
\(\Leftrightarrow\left(x^2-4\right)\left(x+2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2\left(x-2\right)P=\left(x-2\right)\left(x-1\right)Q\)
\(\Leftrightarrow\)\(\left(x+2\right)^2P=\left(x-1\right)Q\)
\(\Leftrightarrow P=x-1\)
\(Q=\left(x+2\right)^2=x^2+4x+4\)
b)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
Đưa về M = x − 1 ( x + 3 ) 2 N . Chọn N = ( x + 3 ) 2 Þ M = x - 1.
x + 2 . P x 2 - 1 = x - 2 . Q x 2 - 2 x + 1
⇒ x + 2 . P . x 2 - 2 x + 1 = x 2 - 1 x - 2 . Q
Hay x + 2 x - 1 2 . P = x - 1 x + 1 x - 2 . Q
Chọn P = (x – 2)(x + 1) = x 2 - x - 2 thì Q = (x + 2)(x – 1) = x 2 + x - 2
