a) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-1\end{matrix}\right.\)
Ta có:
\(A=\left(\frac{1}{x}+\frac{x}{1+x}\right):\frac{x}{x^2+x}\)
\(A=\left[\frac{1+x}{x\left(1+x\right)}+\frac{x^2}{x\left(1+x\right)}\right]:\frac{x}{x\left(1+x\right)}\)
\(A=\frac{x^2+x+1}{x\left(1+x\right)}:\frac{1}{1+x}\)
\(A=\frac{x^2+x+1}{x\left(1+x\right)}.\left(1+x\right)\)
\(A=\frac{x^2+x+1}{x}\)
\(A=x+\frac{1}{x}+1\)
b) Để A = 3 \(\Leftrightarrow x+\frac{1}{x}+1=3\)
\(\Rightarrow x+\frac{1}{x}=3-1=2\)
\(\Rightarrow\frac{x^2+1}{x}=3\)
\(\Rightarrow x^2+1=3x\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow\left(x^2-2x.\frac{3}{2}+\frac{9}{4}\right)-\frac{9}{4}+1=0\)
\(\Rightarrow\left(x-\frac{3}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2=0\)
\(\Rightarrow\left(x-\frac{3}{2}-\frac{\sqrt{5}}{2}\right)\left(x-\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=0\)
\(\Rightarrow\left(x-\frac{3+\sqrt{5}}{2}\right)\left(x-\frac{3-\sqrt{5}}{2}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{matrix}\right.\) ( TMĐK )
c) Để A > 3 \(\Leftrightarrow x+\frac{1}{x}+1>3\)
\(\Rightarrow x+\frac{1}{x}>3-1=2\)
\(\Rightarrow\frac{x^2+1}{x}>2\)
\(\Rightarrow x^2+1>2x\)
\(\Rightarrow x^2-2x+1>0\)
\(\Rightarrow\left(x-1\right)^2>0\)
\(\Rightarrow x-1>0\)
\(\Rightarrow x>1\)
Vậy x > 1 thì A > 3
