$Fe_2O_3 + 3H_2SO_4 \to Fe_2(SO_4)_3 + 3H_2O$

Theo PTHH:  $n_{H_2O} = n_{H_2SO_4} = 0,5.0,1 = 0,05(mol)$
Bảo toàn khối lượng : 

$m_{muối} = 2,81 + 0,05.98 - 0,05.2 = 7,61(gam)$

a: \(A=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)

b: Ta có: \(x^2-2x=0\)

=>x=2

Thay x=2 vào A, ta được:

\(A=\dfrac{2+2}{2+1}=\dfrac{4}{3}\)

(a)

\(A=\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{2x^2-2x-1}{x^2+x}\\ =\dfrac{2x}{x+1}+\dfrac{x-1}{x}-\dfrac{2x^2-2x-1}{x\left(x+1\right)}=\dfrac{2x^2}{x\left(x+1\right)}+\dfrac{x^2-1}{x\left(x+1\right)}-\dfrac{2x^2-2x-1}{x\left(x-1\right)}\)

\(=\dfrac{2x^2+x^2-1-2x^2+2x+1}{x\left(x+1\right)}=\dfrac{x^2+2x+1}{x\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x+1}{x}\)

(b)

\(x^2-2x=0\\ x\left(x-2\right)=0\)

=>x=0 hoặc x=2 mà đk x khác 0 nên thay x=2 vào bt A , ta có:

\(\dfrac{x+1}{x}=\dfrac{2+1}{2}=\dfrac{3}{2}\)

1.He doesn't work in the city anymore
2.They go to work on foot
3.Mai goes to her English class by bike
4.My bike is different from your bike
5.Lan's style is not the same as Minh's

24 B

25 C

26 B

27 C

28 A

29 D

30 C

31 A

32 C

33 B

34 B

35 D

36 C

37 C

38 B

39 C

 1.A     2.B       3. D      4. C      5.B       6. A      7. D      8. C     9. D     10. B

11 B   12  D   13 C   14 A    15 C    16 A  17 D    18 B   19 B    20 C

21 A

22 A

1A 2B 3D 4D 5B 6A 7D 8C 9D 10B 11B 12C 13C 14A 15D 16A 17D 18B 19B 20C 21A 22A