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a. Để P được xđ thì MT phải khác 0.
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9\ne0\\x^2+3x\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(x+3\right)\ne0\\x\left(x+3\right)\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne0\end{matrix}\right.\)
b. \(P=\left(\dfrac{x+9}{x^2-9}-\dfrac{3}{x^2+3x}\right).\dfrac{x-3}{x+3}\)
\(P=\left(\dfrac{x+9}{\left(x-3\right)\left(x+3\right)}-\dfrac{3}{x\left(x+3\right)}\right).\dfrac{x-3}{x+3}\)
\(P=\left(\dfrac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\dfrac{3\left(x-3\right)}{x\left(x+3\right)\left(x-3\right)}\right).\dfrac{x-3}{x+3}\)
\(P=\dfrac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)
\(P=\dfrac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+3}\)
\(P=\dfrac{1}{x}\)
\(A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left[\frac{-\left(x-3\right)}{x+3}.\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left[\frac{-\left(x-3\right)\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)^2}+\frac{x}{x+3}\right].\frac{x+3}{3x^2}\)
\(=\left(-1+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)
\(=\frac{-x-3+x}{x+3}.\frac{x+3}{3x^2}=\frac{-3}{x+3}.\frac{x+3}{3x^2}=\frac{-1}{x^2}\)
b ) Để \(A=-\frac{1}{x^2}< 0\forall x\ne0\)
Vậy \(x\ne0\) thì \(A< 0\)
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)