Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, ĐKXĐ: \(\hept{\begin{cases}x^3+1\ne0\\x^9+x^7-3x^2-3\ne0\\x^2+1\ne0\end{cases}}\)
b, \(Q=\left[\left(x^4-x+\frac{x-3}{x^3+1}\right).\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\frac{\left(x^3+1\right)\left(x^4-x\right)+x-3}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\left[\left(x^7-3\right).\frac{\left(x-1\right)}{\left(x^7-3\right)\left(x^2+1\right)}+1-\frac{2\left(x+6\right)}{x^2+1}\right]\)
\(Q=\frac{x-1+x^2+1-2x-12}{x^2+1}\)
\(Q=\frac{\left(x-4\right)\left(x+3\right)}{x^2+1}\)
\(Q=x^2+y^2=\left(x+y\right)^2-2xy=\left(-1\right)^2-2\left(-6\right)=13\\ P=x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\\ P=\left(-1\right)^3-3\left(-6\right)\left(-1\right)=-1-18=-19\)
\(P=\left(x+y\right)^2-2xy=\left(-1\right)^2-2\cdot\left(-6\right)=1+12=13\)
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
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)\(\dfrac{\left(x+2\right)P}{x^2-1}=\dfrac{\left(x-2\right)Q}{x^2-2x+1}\)
\(\Leftrightarrow\left(x-1\right)^2\left(x+2\right)P=\left(x+1\right)\left(x-1\right)\left(x-2\right)Q\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)P=\left(x+1\right)\left(x-2\right)Q\)
\(\Leftrightarrow P=\left(x+1\right)\left(x-2\right)=x^2-x-2\)
\(Q=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
