Áp dụng bất đửng thức cô si cho các cặp số dương ta có:

\(\left\{{}\begin{matrix}ab+\dfrac{a}{b}\ge2\sqrt{ab\times\dfrac{a}{b}}=2a\\ab+\dfrac{b}{a}\ge2\sqrt{ab\times\dfrac{b}{a}}=2b\\\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\times\dfrac{b}{a}}=2\end{matrix}\right.\)

cộng theo vế 3 bđt trên ta được

\(2\left(ab+\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2\left(a+b+1\right)\)

\(\Leftrightarrow ab+\dfrac{a}{b}+\dfrac{b}{a}\ge a+b+1\)

a,PTHH:

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

_x_________________x_____x

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

_y__________________0,5y________1,5y

b, \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

ta có hệ: \(\left\{{}\begin{matrix}24x+27y=8,7\\x+1,5y=0,4\end{matrix}\right.\rightarrow\left\{{}\begin{matrix}x=0.25\left(mol\right)\\y=0.1\left(mol\right)\end{matrix}\right.\)

do đó: \(m_{Mg}=0.25\times24=6\left(g\right)\)

Vậy :\(\%m_{Mg}=\dfrac{6}{8,7}\times100\approx68,97\%\)

\(\%m_{Al}=100\%-68,97\%=31,03\%\)

c, \(m=m_{MgSO_4}+m_{Al_2\left(SO_4\right)_3}=0,25\times120+0,05\times342=47,1\left(g\right)\)

ta có \(d_{A\O_2 }=0.25\Rightarrow M_A=0.25\times32=8\)

a, Dùng đường chéo ta có

\(\Rightarrow\dfrac{n_{H_2}}{n_{O_2}}=\dfrac{24}{6}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{4}{5}\times100=80\%\\\%V_{O_2}=\dfrac{1}{5}\times100=20\%\end{matrix}\right.\)

b.\(V_A=\dfrac{11.2}{22.4}=0.5\Rightarrow\left\{{}\begin{matrix}V_{H_2}=0.4\left(mol\right)\\V_{O_2}=0,1\left(mol\right)\end{matrix}\right.\)

PTHH: \(2H_2+O_2\rightarrow2H_2O\)

Trước pu 0.4 0.1

Pu 0.2 0.1

Sau pu 0.2 0 0.2

\(\Rightarrow m_{H_2O}=0.2\times18=3,6\left(g\right)\)

ta có: 18km/h=5m/s

54km/h=15m/s

lại có:\(v_1^2-v_0^2=2aS\Rightarrow a=\frac{v_1^2-v_0^2}{2S}=\frac{18^2-15^2}{2\times100}=0,495\left(m\s^2 \right)\)

Dạng 4

1)\(4FeS_2+11O_2\rightarrow2Fe_2O_3+8SO_2\)

2)\(2FeS+9KNO_3\rightarrow9KNO_2+Fe_2O_3+2SO_3\)

3)\(FeS_2+18HNO_3\rightarrow Fe\left(NO_3\right)_3+2H_2SO_4+15NO_2+7H_2O\)

4)\(FeS_2+5HNO_3+3HCl\rightarrow FeCl_3+2H_2SO_4+5NO+2H_2O\)

5)\(3As_2S_3+28HNO_3+4H_2O\rightarrow6H_3AsO_4+9H_2SO_4+28NO\)

6)\(2CrI_3+27Cl_2+64KOH\rightarrow2K_2SO_4+6KIO_4+54KCl+32H_2O\)

7)\(4CuFeS_2+9O_2\rightarrow2Cu_2S+6SO_2+2Fe_2O_3\)

8)\(2FeS+10H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+9SO_2+10H_2O\)

10)\(3Fe_3C+40HNO_3\rightarrow9Fe\left(NO_3\right)_3+3CO_2+13NO+20H_2O\)

chủ ngữ là no woman thì động từ chia ngôi mấy

hệ hả

cho diện tích rồi còn tính chi nữa

theo đề bài ta có:\(\frac{a}{c}=\frac{c}{b}\Rightarrow c^2=ab\)

ta có:

\(\frac{a^2+c^2}{b^2+c^2}=\frac{a^2+ab}{b^2+ab}=\frac{a\left(a+b\right)}{b\left(a+b\right)}=\frac{a}{b}\left(đpcm\right)\)

\(4Na+O_2\rightarrow2Na_2O\)

\(2KClO_3\rightarrow2KCl+3O_2\)