71.

\(\left\{{}\begin{matrix}BB'\perp\left(ABCD\right)\\BB'\in\left(ABB'A'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(ABB'A'\right)\)

74.

\(\left\{{}\begin{matrix}DD'\perp\left(ABCD\right)\\DD'\in\left(CDD'C'\right)\end{matrix}\right.\) \(\Rightarrow\left(ABCD\right)\perp\left(CDD'C'\right)\)

Đề bài thiếu thì phải

32, A

33,B

34,C

31B

32C

33B

34 underline "they"?

Câu 92:

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ ZnO+2HCl\to ZnCl_2+H_2O\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1(mol)\\ \Rightarrow n_{Zn}=0,1(mol)\\ \Rightarrow m_{Zn}=0,1.65=6,5(g)\\ \Rightarrow \%_{Zn}=\dfrac{6,5}{14,6}.100\%\approx44,52\%\\ \Rightarrow \%_{ZnO}=100\%-44,52\%=55,48\%\\ b,m_{ZnO}=14,6-6,5=8,1(g)\\ \Rightarrow n_{ZnO}=\dfrac{8,1}{81}=0,1(mol)\\ \Rightarrow \Sigma n_{HCl}=2n_{Zn}+2n_{ZnO}=0,4(mol)\\ \Rightarrow V_{dd_{HCl}}=\dfrac{0,4}{0,5}=0,8(mol)\)

Câu 93:

\(n_{H_2}=\dfrac{16,8}{22,4}=0,75(mol)\\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2\\ \Rightarrow n_{Fe}=n_{H_2}=0,75(mol)\\ \Rightarrow m_{Fe}=0,75.56=42(g)\\ b,n_{H_2SO_4}=n_{H_2}=0,75(mol)\\ \Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,75}{0,25}=3M\\ c,n_{FeSO_4}=0,75(mol)\\ \Rightarrow m_{CT_{FeSO_4}}=0,75.152=114(g)\\ V_{dd_{FeSO_4}}=V_{dd_{H_2SO_4}}=250(ml)\\ \Rightarrow m_{dd_{FeSO_4}}=250.1,1=275(g)\\ \Rightarrow C\%_{FeSO_4}=\dfrac{114}{275}.100\%\approx41,45\%\)

\(d,m_{FeSO_4.5H_2O}=242.0,75=181,5(g)\)

\(\dfrac{x}{y}=\dfrac{5}{2}\) ⇒\(\dfrac{x}{5}=\dfrac{y}{2}\)

\(\dfrac{y}{z}=\dfrac{1}{3}\) ⇒\(\dfrac{y}{1}=\dfrac{z}{3}\) ⇒\(\dfrac{y}{2}=\dfrac{z}{6}\)

⇒\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{6}\) ⇒\(\dfrac{x^2}{25}=\dfrac{y^2}{4}=\dfrac{2z^2}{72}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x^2}{25}=\dfrac{y^2}{4}=\dfrac{2z^2}{72}=\dfrac{x^2-y^2+2z^2}{25-4+72}=\dfrac{372}{93}=4\)

⇒\(\left\{{}\begin{matrix}x=4.5=20\\y=4.2=8\\z=4.6=24\end{matrix}\right.\)

Câu 8: 

#include <bits/stdc++.h>

using namespace std;

long long a[250],n,i;

int main()

{

cin>>n;

for (i=1; i<=n; i++) cin>>a[i];

sort(a+1,a+n+1);

for (i=1; i<=n; i++) cout<<a[i]<<" ";

return 0;

}