\(C+O_2\underrightarrow{t^o}CO_2\)

\(\text{Theo định luật bảo toàn khối lượng}\)

\(m_C+m_{O_2}=m_{CO_2}\)

\(\Leftrightarrow1,2g+4,8g=m_{CO_2}\)

\(\Rightarrow m_{CO_2}=6\left(g\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

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

\(0,02mol\)   \(0,02mol\)         \(0,02mol\)

\(n_{Zn}=\dfrac{m}{M}=\dfrac{1,95}{65}=0,03\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{1,57}{98}\approx0,02\left(mol\right)\)

\(\text{Ta thấy }Zn\text{ dư,}H_2SO_4\text{ phản ứng hết}\)

\(n_{Zn_{dư}}=0,03-0,02=0,01\left(mol\right)\)

\(m_{Zn_{dư}}=n.M=0,01.65=0,65\left(mol\right)\)

\(V_{H_2}=n.22,4=0,02.22,4=0,448\left(l\right)\)

\(\text{Gọi M là kim loại hóa trị II}\)

\(M+2HCl\rightarrow MCl_2+H_2\)

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

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

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(M_M=\dfrac{m}{n}=\dfrac{4,8}{0,2}=24\)

\(\text{Vậy kim loại hóa trị II là Magie(Mg)}\)

\(3Fe+2O_2\underrightarrow{t^o}Fe_3O_4\)

        \(2mol\)    \(1mol\)

      \(0,02mol\)    \(0,01mol\)

\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{2,32}{232}=0,01\left(mol\right)\)

\(V_{O_2}=n.22,4=0,02.22,4=0,048\left(l\right)\)

\(x\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=9\end{matrix}\right.\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{0;-3;9\right\}\)

\(a)7x+9=30\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=3\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{3\right\}\)

\(b)3\left(x-1\right)-2=4\)

\(\Leftrightarrow3x-3-2=4\)

\(\Leftrightarrow3x-5=4\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{3\right\}\)

\(c)\dfrac{x+2}{4}=\dfrac{x-1}{6}+\dfrac{3-x}{3}\)

\(\Leftrightarrow\dfrac{3\left(x+2\right)}{12}=\dfrac{2\left(x-1\right)}{12}+\dfrac{4\left(3-x\right)}{12}\)

\(\Leftrightarrow3x+6=2x-2+12-4x\)

\(\Leftrightarrow3x+6=10-2x\)

\(\Leftrightarrow3x+6-10+2x=0\)

\(\Leftrightarrow5x-4=0\)

\(\Leftrightarrow5x=4\)

\(\Leftrightarrow x=\dfrac{4}{5}\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{4}{5}\right\}\)

\(d)\dfrac{x+1}{x}-\dfrac{3x}{x\left(x+4\right)}=\dfrac{x-2}{x+4}\text{ĐKXĐ:}x\ne0;x\ne-4\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+4\right)}{x\left(x+4\right)}-\dfrac{3x}{x\left(x+4\right)}=\dfrac{x\left(x-2\right)}{x\left(x+4\right)}\)

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

\(\Leftrightarrow x^2+2x+4=x^2-2x\)

\(\Leftrightarrow x^2+2x+4-x^2+2x=0\)

\(\Leftrightarrow4x+4=0\)

\(\Leftrightarrow4x=-4\)

\(\Leftrightarrow x=-1\left(TM\text{Đ}KX\text{Đ}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-1\right\}\)

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)

\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)

\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)

\(\Leftrightarrow\left(x-2014\right).A=0\)

\(\text{Vì A }\ne0\)

\(\Rightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)

\(\text{Xét }\Delta ABC\text{ có:}\)

\(BD\text{ là tia phân giác}\left(gt\right)\)

\(\Rightarrow\dfrac{BC}{CD}=\dfrac{AB}{AD}\Rightarrow\dfrac{5}{2,5}=\dfrac{4}{y}\Rightarrow y=\dfrac{2,5.4}{5}=2\left(cm\right)\)

\(\text{Xét }\Delta EDF\text{ có:}\)

\(DI\text{ là đường phân giác(gt)}\)

\(\Rightarrow\dfrac{ED}{EI}=\dfrac{DF}{IF}\Rightarrow\dfrac{3,5}{y}=\dfrac{6}{3}\Rightarrow y=\dfrac{3,5.3}{6}=1,75\)

\(\text{Xét }\Delta ABC\text{ có:}\)

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

\(\Rightarrow\dfrac{AM}{BM}=\dfrac{AN}{CN}\left(\text{định lí Ta-lét}\right)\)

\(\Rightarrow\dfrac{6}{3}=\dfrac{4}{x}\Rightarrow x=\dfrac{3.4}{6}=2\)