\(DF=DN+NF=4+8=12\left(cm\right)\)

\(\text{Xét }\Delta DMN\text{ và }\Delta DEF\text{ có:}\)

\(MN//EF\left(gt\right)\)

\(\Rightarrow\dfrac{DN}{DF}=\dfrac{MN}{EF}\left(\text{hệ quả Ta lét}\right)\)

\(\Rightarrow\dfrac{4}{12}=\dfrac{3}{x}\)

\(\Rightarrow x=\dfrac{12.3}{4}=9\left(cm\right)\)

\(\dfrac{x}{30}=\dfrac{x}{40}+\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{4x}{120}=\dfrac{3x}{120}+\dfrac{90}{120}\)

\(\Leftrightarrow4x=3x+90\)

\(\Leftrightarrow4x-3x-90=0\)

\(\Leftrightarrow x-90=0\)

\(\Leftrightarrow x=90\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{90\right\}\)

\(n_{SO_2}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

\(V_{SO_2}=n.22,4=0,1.22,4=2,24\left(l\right)\)

\(\Rightarrow A\)

\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\text{ĐKXĐ:}x\ne3;-1;\)

\(\Leftrightarrow\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2.2x}{2\left(x+1\right)\left(x-3\right)}MTC:2\left(x+1\right)\left(x-3\right)\)

\(\Rightarrow x^2+x+x^2-3x=4x\)

\(\Leftrightarrow2x^2-2x=4x\)

\(\Leftrightarrow2x^2-2x-4x=0\)

\(\Leftrightarrow2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\left(\text{loại}\right)\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{0\right\}\)

\(3Fe+2O_2\rightarrow Fe_3O_4\)

\(3mol\)      \(2mol\)     \(1mol\)

\(0,3mol\)   \(0,2mol\)   \(0,1mol\)

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

\(m_{Fe}=n.M=0,3.56=16,8\left(g\right)\)

\(V_{O_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)

\(2mol\)                                              \(1mol\)'

\(0,3mol\)                                           \(0,15mol\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

\(m_{KMnO_4}=n.M=0,3.158=47,4\left(g\right)\)

\(19x-7=0\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)

\(a)3Fe+2O_2\rightarrow Fe_3O_4\)

\(3mol\)   \(2mol\)        \(1mol\)

\(0,3mol\)  \(0,2mol\)   \(0,1mol\)

\(b)n_{Fe}=\dfrac{m}{M}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

\(\text{Ta thấy }O_2\text{ dư,}Fe\text{ phản ứng hết}\)

\(c)m_{Fe_3O_4}=n.M=0,1.232=23,2\left(g\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(1mol\)     \(2mol\)                 \(1mol\)

\(0,2mol\)    \(0,4mol\)          \(0,2mol\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

\(V_{H_2}=n.22,4=0,2.22,4=4,48\left(l\right)\)

\(m_{HCl}=n.M=0,4.=14,6\left(g\right)\)

\(\Rightarrow12x=9\left(x+1\right)\)

\(\text{Gọi thời gian lúc đi là x(h)}\)

\(\left(\text{ĐK}:x>0\right)\)

\(\text{Thời gian lúc về là x+1(h) }\)

\(\text{Quãng đường lúc đi là 12x(km)}\)

\(\text{Quãng đường lúc về là 9(x+1)(km)}\)

\(\text{Theo đề ta có:}12x=9\left(x+1\right)\)

                    \(\Leftrightarrow12x=9x+9\)

                    \(\Leftrightarrow12x-9x-9=0\)

                    \(\Leftrightarrow3x-9=0\)

                    \(\Leftrightarrow3x=9\)

                    \(\Leftrightarrow x=3\)

\(\text{Vậy quãng đường AB dài là:12.3=36(km)}\)