c) Để P=3 thì \(\dfrac{x+1}{2x}=3\)
\(\Leftrightarrow x+1=6x\)
\(\Leftrightarrow x-6x=-1\)
\(\Leftrightarrow-5x=-1\)
hay \(x=\dfrac{1}{5}\)(thỏa ĐK)
Vậy: Để P=3 thì \(x=\dfrac{1}{5}\)
a) Ta có: \(A=\dfrac{1}{x^2+x}+\dfrac{1}{x+1}\)
\(=\dfrac{1}{x\left(x+1\right)}+\dfrac{x}{x\left(x+1\right)}\)
\(=\dfrac{x+1}{x\left(x+1\right)}=\dfrac{1}{x}\)
b) Ta có: \(P=\dfrac{A}{B}\)
\(=\dfrac{1}{x}:\dfrac{2}{x+1}\)
\(=\dfrac{1}{x}\cdot\dfrac{x+1}{2}\)
\(=\dfrac{x+1}{2x}\)
e) Ta có: \(P>\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x+1}{2x}-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{x+1}{2x}-\dfrac{x}{2x}>0\)
\(\Leftrightarrow\dfrac{1}{2x}>0\)
\(\Leftrightarrow2x>0\)
hay x>0
Kết hợp ĐKXĐ, ta được: x>0
Vậy: Để \(P>\dfrac{1}{2}\) thì x>0