Bài 2:
a: \(P=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)
b: Để P là số nguyên dương thì \(\left\{{}\begin{matrix}\dfrac{3x}{x-3}>0\\3x⋮x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{x-3}>0\\3x-9+9⋮x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\\9⋮x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\\x-3\in\left\{1;-1;3;-3;9;-9\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< 0\end{matrix}\right.\\x\in\left\{4;2;6;0;12;-6\right\}\end{matrix}\right.\)
=>\(x\in\left\{4;6;12;-6\right\}\)
c: \(Q=P\cdot\dfrac{x-3}{5}=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{5}=\dfrac{3x}{5}\)
Để \(Q=\dfrac{1}{5}\) thì \(\dfrac{3x}{5}=\dfrac{1}{5}\)
=>3x=1
=>\(x=\dfrac{1}{3}\left(nhận\right)\)
Bài 1:
a: Thay x=2 vào A, ta được:
\(A=\dfrac{2^2-2}{2\cdot2+1}=\dfrac{2}{5}\)
b: \(P=A\cdot B\)
\(=\dfrac{x\left(x-1\right)}{2x+1}\cdot\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{x\left(x-1\right)}{2x+1}\cdot\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)}{2x+1}\cdot\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{x}{x+1}\)
c: Đặt \(Q=3\cdot P=3\cdot\dfrac{x}{x+1}=\dfrac{3x}{x+1}\)
Để Q là số nguyên thì \(3x⋮x+1\)
=>\(3x+3-3⋮x+1\)
=>\(-3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
