7C 8D

Câu 7: B

Câu 8: C

Câu 7:

a: \(18+\dfrac{9}{3}=18+3=21\)

b: \(\dfrac{4}{9}-\dfrac{5}{18}=\dfrac{8}{18}-\dfrac{5}{18}=\dfrac{3}{18}=\dfrac{1}{6}\)

c: \(\dfrac{3}{5}:\dfrac{5}{7}=\dfrac{3}{5}\cdot\dfrac{7}{5}=\dfrac{21}{25}\)

d: \(\dfrac{4}{7}\cdot\dfrac{2}{7}=\dfrac{4\cdot2}{7\cdot7}=\dfrac{8}{49}\)

Câu 8:

a: \(\dfrac{5}{4}\cdot x=\dfrac{3}{8}+\dfrac{5}{12}\)

=>\(x\cdot\dfrac{5}{4}=\dfrac{9+10}{24}=\dfrac{19}{24}\)

=>\(x=\dfrac{19}{24}:\dfrac{5}{4}=\dfrac{19}{24}\cdot\dfrac{4}{5}=\dfrac{19}{6\cdot5}=\dfrac{19}{30}\)

b: \(3\cdot\dfrac{7}{10}+\dfrac{7}{10}\cdot5+2\cdot\dfrac{7}{10}\)

\(=\dfrac{7}{10}\left(3+5+2\right)\)

\(=\dfrac{7}{10}\cdot10=7\)

đề bài?

7/

4FeS2 + 11O2 → 2Fe2O3 + 8SO2

FexOy +(x-y)CO ----> xFeO +(x - y)CO2

Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

2KClO3 ⟶ 2KCl + 3O2

2Fe(OH)3 ⟶ Fe2O3 + 3H2O

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

8. \(4M+nO_2-^{t^o}\rightarrow2M_2O_n\)

Theo PT : \(n_M=2n_{M_2O_n}\)

=> \(\dfrac{5,4}{M}=2.\dfrac{10,2}{2M+16n}\)

Chạy nghiệm n=1,2,3

n=1 => M= 9 (loại)

n=2 => M= 18 (loại)

n=3 => M= 27 (chọn)

Vậy M là Al

Câu 7 : H/s x/đ \(\Leftrightarrow x^2+5x+6\ne0\Leftrightarrow\left(x+2\right)\left(x+3\right)\ne0\)  \(\Leftrightarrow x\ne-2;-3\)

Chọn B 

Câu 8 : A