why can't use only one way function to construct a PRG, and don't use the hard core predicate? The one-way function can expose some of the input and stay one-way, but still, that doesn't give the input.


As I read, we can't just construct a PRG from a OWF, but we need to use the hard core predicate of the OWF. Why is that?
Assuming $|f(s)| &gt; |s|$, why is constructing a PRG in the following way wrong? $G(s) = f(s)$
Suppose $f$ is a permutation function. If $f$ gets a random $x$, then is $f(x)$ also random?
The first one is really simple, you already said it yourself:


  as i read , we can't just construct a PRG from just OWF , but we need to use the hard core predicate of the OWF, why is that?


I guess you're familiar with the formal statements, but informally speaking we have:


A function $f:\{0,1\}^* \rightarrow \{0,1\}^*$ is one-way, if we give the adversary $y = f(x)$ and the length $|x|$, and he has a negligible probability of finding $x'$, s.t. $f(x') = y$.
a PRG is a function $g:\{0,1\}^n \rightarrow \{0,1\}^m$, where the output is indisinguishable from $U^m$, denoting the uniform distribution over $m$ bits, and $m &gt; n$


Now consider the following construction: $f'(x) = f(x)|1$, with $f$ being a OWF. Just concatenating a $1$ doesn't change much w.r.t. the security property of an OWF. And in fact, any adversary able to break $f(x)$ can easily be turned into an adversary for $f'(x)$ and vice versa. So $f'$ is a OWF iff $f$ is a OWF.

But it is quite obvious, that $f'$ is not a PRG, because the output of $f'$ can easily be distinguished from a uniform distribution, by just looking at the lowest ordered bit (always $1$ vs uniform).

But as you stated yourself: With a hardcore predicate, you can get a length extension. Maybe these lecture notes make it more clear how to achieve that (and why it's necessary). Informally speaking: A OWF does not need to be impossible to invert for all parts of its output - while a PRG needs to be indistinguishable from randomness over all its output. 


  why constructing prg in the following way is wrong G(s) = f(s) assuming |f(s)| > |s|?


If we specify $f(x):\{0,1\}^n \rightarrow \{0,1\}^n$ a length preserving OWF, and use the construction above, we get $f':\{0,1\}^n \rightarrow \{0,1\}^{n+1}$, which is also a OWF. This fulfills the length criteria of a PRG, but it is distinguishable. Being a OWF is not sufficient to be a PRG. 


  f is a permutation function , if f gets a random x then is f(x) also random?


A permutation is a bijection from a set, e.g. $\{0,1\}^n$ into itself. This means, there are no collisions and every element has a single preimage. Thus, if we consider an input drawn uniformly at random from $\{0,1\}^n$, then the output is also uniformly distributed over $\{0,1\}^n$ - because the permutation has a 1-to-1 relationship between inputs and outputs. But that is also fullfilled for example by $p(x) = x + 1 \mod n$ over $\mathbb{Z}_n$, and I don't think that's what you meant. But this is entirely unrelated to the concept of one-way functions and PRGs.