b) \(2x=\left(3,8\right)^3:\left(-3,8\right)^3\)

\(\Rightarrow2x=-1\)

⇒ \(x=\left(-1\right):2\)

⇒ \(x=-\frac{1}{2}\)

Vậy \(x=-\frac{1}{2}.\)

Chúc bạn học tốt!

\(2x:6=5:3\)                              \(-\frac{2}{x}=-\frac{x}{8}\)

\(\Rightarrow2x:6=\frac{5}{3}\)                          \(\Rightarrow-2\times8=x\times\left(-x\right)\)

\(\Rightarrow2x=\frac{5}{3}\times6\)                      \(\Rightarrow-16=\left(-x\right)^2\)

\(\Rightarrow2x=10\)                               \(\Rightarrow\left(-4\right)^2=\left(-x\right)^2\)

\(\Rightarrow x=5\)                                    \(\Rightarrow x=-4\)

Có \(\dfrac{m_O}{m_N}=\dfrac{16.n_O}{14.n_N}=\dfrac{80}{21}\)

=> \(\dfrac{n_O}{n_N}=\dfrac{10}{3}\)

\(n_{HCl}=0,03.1=0,03\left(mol\right)\)

=> \(n_N=0,03\left(mol\right)\)

=> \(n_O=0,1\left(mol\right)\)

Bảo toàn N: \(n_{N_2}=0,015\left(mol\right)\)

Gọi \(\left\{{}\begin{matrix}n_{CO_2}=a\left(mol\right)\\n_{H_2O}=b\left(mol\right)\end{matrix}\right.\)

\(n_{O_2}=\dfrac{3,192}{22,4}=0,1425\left(mol\right)\)

Bảo toàn O: 0,1 + 0,1425.2 = 2a + b

=> 2a + b = 0,385

Bảo toàn khối lượng: \(44a+18b+0,015.28=3,83+0,1425.32\)

=> 44a + 18b = 7,97

=> a = 0,13 ; b = 0,125

\(n_{CaCO_3}=n_{CO_2}=0,13\left(mol\right)\)

=> \(m_{CaCO_3}=0,13.100=13\left(g\right)\)

3,8 : 2x = 1/4 : 2 + 2/3

3,8 : 2x = 19/24

        2x = 24/5

          x = 12/5

       \(3,8:2x=\frac{1}{4}:\frac{8}{3}\)

      \(3,8:2x=\frac{1}{4}\cdot\frac{3}{8}\)

      \(3,8:2x=\frac{3}{32}\)

      \(2x=3,8:\frac{3}{32}\)

      \(2x=\frac{19}{5}\cdot\frac{32}{3}\)

b) \(3,8:\left(2x\right)=\frac{1}{4}:\frac{8}{3}\)

= \(3,8:\left(2x\right)=\frac{3}{32}\)

\(\Rightarrow2x=3,8:\frac{3}{32}=\frac{608}{15}\)

\(\Rightarrow x\frac{608}{15}:2=\frac{304}{15}\)

\(\Rightarrow\frac{304}{15}\)