a) \(3Fe+2O_2-t^o->Fe_3O_4\)

b) \(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)

Theo pthh : \(n_{Fe_3O_4}=\frac{1}{3}n_{Fe_3O_4}=\frac{0,1}{3}\left(mol\right)\)

=> \(m_{Fe_3O_4}=232\cdot\frac{0,1}{3}\approx7,73\left(g\right)\)

c) Theo pthh : \(n_{O2\left(pứ\right)}=\frac{2}{3}n_{Fe}=\frac{0,2}{3}\left(mol\right)\)

=> \(n_{O2\left(can.dung\right)}=\frac{0,2}{3}\div100\cdot120=0,08\left(mol\right)\)

=> \(V_{O2\left(can.dung\right)}=0,08\cdot22,4=1,792\left(l\right)\)

a, \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)

\(\Leftrightarrow\frac{6x-3}{15}-\frac{5x-10}{15}=\frac{x+7}{15}\)

Khử mẫu : \(6x-3-5x+10=x+7\)

\(\Leftrightarrow7+x=x+7\Leftrightarrow0=0\)( vip :') 

d, \(\frac{x+1}{2019}+\frac{x+2}{2018}=\frac{x+3}{2017}+\frac{x+4}{2016}\)

\(\Leftrightarrow\frac{x+1}{2019}+1+\frac{x+2}{2018}+1=\frac{x+3}{2017}+1+\frac{x+4}{2016}+1\)

\(\Leftrightarrow\frac{x+2020}{2019}+\frac{x+2020}{2018}-\frac{x+2020}{2017}-\frac{x+2020}{2016}=0\)

\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{2019}+\frac{1}{2018}-\frac{1}{2017}-\frac{1}{2016}\ne0\right)=0\)

\(\Leftrightarrow x=-2020\)

a,\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)

\(\Leftrightarrow\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}=\frac{x+7}{15}\)

\(\Leftrightarrow6x-3-5x+10=x+7\)

\(\Leftrightarrow6x-3-5x+10-x-7=0\)

\(\Leftrightarrow\left(6x-5x-x\right)-\left(3-10+7\right)=0\)

\(\Leftrightarrow0=0\)

Vậy....

program sosanh;

uses crt;

var a,b,c : real;

begin

      readln(a,b);

      c:= a-b;

      if c>0 then writeln(' a lon hon b');

      if c<0 then writeln('a nho hon b');

      if c=0 then writeln('a bang b');

      readln;

end.

Dốt tin nên chỉ làm dc vậy, đúng thì 5 sao nhé

B