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

\(\left(x+\frac{3}{5}\right).\left(\frac{-4}{3}-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{5}=0\\\frac{-4}{3}-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{-4}{3}\end{cases}}}\)

\(\left(x+\frac{3}{5}\right)\left(-\frac{4}{3}-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{3}{5}=0\\-\frac{4}{3}-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{3}{5}\\x=-\frac{4}{3}\end{cases}}}\)

vậy \(x=-\frac{3}{5}\)hoặc \(x=-\frac{4}{3}\).

ta có : \(\left(x+\frac{3}{5}\right)\left(-\frac{4}{3}-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{5}=0\\\frac{-4}{3}-x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{5}\\x=-\frac{4}{3}\end{cases}}\)

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 - 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

Biến đổi ta được:  1 x + 20 . T = 1 2 ⇒ T = x + 20 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)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)

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

\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)

Hì , giải đc rùi nha.

Vì \(x,y\in R\)

\(\Rightarrow\left(x+2\right).\left(y+2\right)=\frac{25}{4}\)

Min \(P=\sqrt{1+x^4}+\sqrt{1+y^4}\)

- Dự đoán \(x=y=\frac{1}{2}\)

- Sử dụng BĐT : \(\frac{x^2}{a}+\frac{y^2}{b}\ge\frac{\left(x+y\right)^2}{a+b}\)    ( Với a,b > 0 )

=>  \(1+x^4=16.\frac{1}{16}+a^4=16.\left(\frac{1}{4}\right)^2+a^2\ge\frac{[16.\frac{1}{4}+a^2]^2}{17}\)

\(=\frac{(a^2+4)^2}{17}\)

=> \(1+y^4\ge\frac{\left(y^2+4\right)^2}{17}\)

=> \(P\ge\frac{x^2+y^2+8}{\sqrt{17}}\)

\(\Leftrightarrow P\sqrt{17}=\frac{1}{5}\left(x^2+y^2\right)+\frac{4}{5}\left(x^2+\frac{1}{4}+y^2+\frac{1}{4}\right)+8-\frac{2}{5}\)

\(\ge\frac{2xy}{5}+\frac{4}{5}\left(x+y\right)+8-\frac{2}{5}=\frac{2}{5}[xy+2\left(x+y\right)]+8-\frac{2}{5}\)

Theo giả thiết \(\left(x+2\right)\left(y+2\right)=\frac{25}{4}\)

\(\Leftrightarrow xy+2\left(x+y\right)=\frac{9}{4}\)

\(\Rightarrow P\sqrt{17}\ge\frac{2}{5}.\frac{9}{4}+8-\frac{2}{5}=\frac{17}{2}\)

\(\Leftrightarrow P\ge\frac{\sqrt{17}}{2}\)

Điểm rơi \(x=y=\frac{1}{2}\)