Đáp án:
$\rm E<F$
Giải thích các bước giải:
$\rm E=\dfrac{n}{n+1}\\\Rightarrow E=\dfrac{n+1-1}{n+1}\\\Rightarrow E=\dfrac{n+1}{n+1}-\dfrac{1}{n+1}\\\Rightarrow E=1-\dfrac{1}{n+1}\\F=\dfrac{n+2}{n+3}\\\Rightarrow F=\dfrac{n+3-1}{n+3}\\\Rightarrow F=\dfrac{n+3}{n+3}-\dfrac{1}{n+3}\\\Rightarrow F=1-\dfrac{1}{n+3}\\\Rightarrow\dfrac{1}{n+1}>\dfrac{1}{n+3}\\\Rightarrow1-\dfrac{1}{n+1}<1-\dfrac{1}{n+3}\\\Rightarrow\dfrac{n}{n+1}<\dfrac{n+2}{n+3}$
Vậy $\rm E<P$.
