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a: \(A=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)
b: Ta có: \(x^2-2x=0\)
=>x=2
Thay x=2 vào A, ta được:
\(A=\dfrac{2+2}{2+1}=\dfrac{4}{3}\)
(a)
\(A=\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{2x^2-2x-1}{x^2+x}\\ =\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{2x^2-2x-1}{x\left(x+1\right)}=\dfrac{2x^2}{x\left(x+1\right)}+\dfrac{x^2-1}{x\left(x+1\right)}-\dfrac{2x^2-2x-1}{x\left(x-1\right)}\)
\(=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x+1}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x+1}{x}\)
(b)
\(x^2-2x=0\\ x\left(x-2\right)=0\)
=>x=0 hoặc x=2 mà đk x khác 0 nên thay x=2 vào bt A , ta có:
\(\dfrac{x+1}{x}=\dfrac{2+1}{2}=\dfrac{3}{2}\)
(a) \(A=\dfrac{3}{x-2}\in Z\)
\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;0;2;4\right\}.\)
(b) \(B=-\dfrac{11}{2x-3}\in Z\)
\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;1;2;7\right\}.\)
(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)
\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=-3\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)
(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)
\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)