e bấm máy nhầm thành 5,4 á a :v

\(M_X=15.M_{H_2}=15.2=30(g/mol)\)

Trong 1 mol X: \(\begin{cases} n_C=\dfrac{30.80\%}{12}=2(mol)\\ n_H=\dfrac{30.20\%}{1}=6(mol) \end{cases}\)

\(\Rightarrow CTHH:C_2H_6\)

\(\Rightarrow x\in\left\{-9;-3;-1;1;3;9\right\}\\ \text{Mà }-2< x\le3\\ \Rightarrow x\in\left\{-1;1;3\right\}\)

\(Fe+CuCl_2\to FeCl_2+Cu\)

Đặt \(n_{CuCl_2}=x(mol)\)

\(\Rightarrow m_{KL tăng}=m_{Cu}-m_{Fe}=3,2(g)\\ \Rightarrow 64x-56x=3,2\\ \Rightarrow x=0,4(mol)\\ \Rightarrow m_{Fe}=0,4.56=22,4(g)\)

Cửa hàng còn lại \(420\times\left(100-65\right)\%=147\left(m\right)\)

Vì \(12345:6=2057\) dư \(3\) nên cần ít nhất \(2057+1=2058\) hộp

Số hs nam là \(36\times\left(100-75\right)\%=9\left(hs\right)\)

\(a,\Rightarrow\left[{}\begin{matrix}x-1=1\\1-x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\\ b,\Rightarrow7^{2x-6}=7^2\Rightarrow2x-6=2\Rightarrow x=4\\ c,\Rightarrow\left(2x-6\right)^7=2^7\Rightarrow2x-6=2\Rightarrow x=4\\ d,\Rightarrow13\cdot x=195\Rightarrow x=15\\ e,\Rightarrow\left(135-7x\right):9=8\Rightarrow135-7x=72\Rightarrow7x=63\Rightarrow x=9\\ f,\Rightarrow338-113+x=158\Rightarrow225+x=158\Rightarrow x=-67\\ g,\Rightarrow3\left(x+2\right):7=30\Rightarrow3\left(x+2\right)=210\Rightarrow x+2=70\Rightarrow x=68\\ h,\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ i,\Rightarrow\left[{}\begin{matrix}x+2=0\\x-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\\ k,\Rightarrow\left(x-140\right):7=27-24=3\\ \Rightarrow x-140=21\Rightarrow x=161\\ l,\Rightarrow x^5=2^5\Rightarrow x=2\\ m,\Rightarrow3^{x-3}=2\cdot3^2+3^2=3^3\Rightarrow x-3=3\Rightarrow x=6\)