=(-37).(-381)-14134

=14097-14134

=-37

=-37

\(=-37+37.382-382.37=-37\)

382*8+2*382

=382*(8+2)

=382*10

=3820

\(-382-\left[125+\left(-382\right)\right]-\left[\left(3210-697\right)+\left(-3210-957\right)\right]\)

\(=-382-125+382-3210+697+3210+957\)

\(=-125-697-957\)

\(=-1779\)

hok tốt!!

Chọn  A. 567 234

a  đúng ko

654

654

576 + 869 + 8967 

= (576 + 869) + 8967

= 1445 + 8967

= 10412 

cảm ơn

\(\frac{\frac{125}{8}+\frac{125}{97}+\frac{125}{576}+\frac{250}{991}}{\frac{25}{8}+\frac{25}{97}+\frac{25}{576}+\frac{50}{991}}\)=\(\frac{250.\left(\frac{1}{8}+\frac{1}{97}+\frac{1}{576}+\frac{1}{991}\right)}{50.\left(\frac{1}{8}+\frac{1}{97}+\frac{1}{576}+\frac{1}{991}\right)}\)=\(\frac{250}{50}\)=5

Nhớ tick

\(\frac{1763}{191}\)

Gọi số mol Al, Cu, Fe là a, b , c

=> 27a + 64b + 56c = 17,5

\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH: 2Al + 6HCl --> 2AlCl₃ + 3H₂

______a------------------------>1,5a

Fe + 2HCl --> FeCl₂ + H₂

c------------------------>c

=>1,5a + c = 0,3

PTHH: 2Al + 3Cl₂ --t^o--> 2AlCl₃
_____a------------------>a

2Fe + 3Cl₂ --t^o--> 2FeCl₃

c------------------->c

Cu + Cl₂ --t^o--> CuCl₂

b---------------->b

=> 133,5a + 135b + 162,5c = 51,225

=> a = 0,1; b = 0,1; c = 0,15

=> \(\%Fe=\dfrac{56.0,15}{17,5}.100\%=48\%\)

=> A