Ai giúp mik với

Định luật II Niu tơn: \(\overrightarrow{F}+\overrightarrow{F_{ms}}=m.\overrightarrow{a}\) 

\(\Rightarrow F-F_{ms}=m.a\)

\(\Rightarrow F-\mu mg=m.a\)

Gia tốc thùng:

\(a=\dfrac{F-\mu mg}{m}=\dfrac{150-0,5\times10\times9,8}{10}=10,1\left(m/s^2\right)\)

\(H_2SO_4.đặc.nóng?\\ n_S=\dfrac{1,6}{32}=0,05mol\\ BTe:2n_{SO_4^{2-}}=2n_{SO_2}+6n_S\\ < =>2n_{SO_4^{2-}}=2.0,2+6.0,05\\ < =>n_{SO_4^{2-}}=0,35mol\\ m_{muối}=7,5+0,35.96=41,1g\)

\(\Delta=\left(2m+1\right)^2-4\left(m^2-1\right)\)

\(=4m^2+4m+1-4m^2+4=4m+5\)

Để phương trình có hai nghiệm phân biệt thì Δ>0

=>4m+5>0

=>4m>-5

=>\(m>-\frac54\)

Theo Vi-et, ta có: \(\begin{cases}x_1+x_2=-\frac{b}{a}=2m+1\\ x_1x_2=\frac{c}{a}=m^2-1\end{cases}\)

Sửa đề: \(A=\left(2x_1-x_2\right)\left(x_1-2x_2\right)\)

\(=2x_1^2-4x_1x_2-x_1x_2+2x_2^2\)

\(=2\left(x_1^2+x_2^2\right)-5x_1x_2\)

\(=2\left\lbrack\left(x_1+x_2\right)^2-2x_1x_2\right\rbrack-5x_1x_2\)

\(=2\left(x_1+x_2\right)^2-9x_1x_2\)

\(=2\left(2m+1\right)^2-9\left(m^2-1\right)\)

\(=2\left(4m^2+4m+1\right)-9m^2+9\)

\(=8m^2+8m+2-9m^2+9=-m^2+8m+11\)

\(=-\left(m^2-8m-11\right)=-\left(m^2-8m+16-27\right)\)

\(=-\left(m-4\right)^2+27\le27\forall m\)

Dấu '=' xảy ra khi m-4=0

=>m=4(nhận)

#include <bits/stdc++.h>
using namespace std;
bool check(long long n)
{
    for(int i=2; i<=sqrt(n); i++)
    {
        if(n%i==0) return false;
    }
    return n>1;
}
signed main()
{
    long long n;
    cin>>n;
    for(int i=1; i<=n; i++)
    {
        long long x;
        cin>>x;
        if(check(x)) cout<<x<<' ';
    }
}