\(A=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+1+y^2-8y+16-17\\ A=\left(x-y+1\right)^2+\left(y-4\right)^2-16\ge17\)

Vậy \(A_{min}=17\leftrightarrow\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Xét các số thực dương \(a,b,c\) thỏa mãn \(abc=a+b+c+2\). CMR:

$$\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\le \dfrac{3}{4}$$

$a)2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2$ (phân hủy)

$b)2Cu+O_2\xrightarrow{t^o}2CuO$ (hóa hợp)

$c)4H_2+Fe_3O_4\xrightarrow{t^o}3Fe+4H_2O$ (thế)

$d)3H_2O+P_2O_5\to 2H_3PO_4$ (hóa hợp)

\(\begin{array} {l} a)\\ n_{Ba(OH)_2}=\dfrac{300.17,1\%}{171}=0,3(mol)\\ n_{H_2SO_4}=\dfrac{100.19,6\%}{98}=0,2(mol)\\ Ba(OH)_2+H_2SO_4\to BaSO_4\downarrow+2H_2O\\ n_{Ba(OH)_2}>n_{H_2SO_4}\to Ba(OH)_2\,dư\\ n_{BaSO_4}=n_{H_2SO_4}=0,2(mol)\\ m_A=m_{BaSO_4}=0,2.233=46,6(g)\\ b)\\ X:Ba(OH)_2\,dư\\ n_{Ba(OH)_2(pứ)}=n_{H_2SO_4}=0,2(mol)\\ Ba(OH)_2+Al_2O_3\to Ba(AlO_2)_2+H_2O\\ n_{Al_2O_3}=n_{Ba(OH)_2(dư)}=0,3-0,2=0,1(mol)\\ m_{Al_2O_3}=0,1.102=10,2(g) \end{array}\)

\(\begin{array} {l} a)\\ n_{NaOH}=\dfrac{8}{40}=0,2(mol)\\ 2NaOH+FeCl_2\to Fe(OH)_2\downarrow+2NaCl\\ n_{FeCl_2}=\dfrac{1}{2}n_{NaOH}=0,1(mol)\\ m_{dd\,FeCl_2}=\dfrac{0,1.127}{25,4\%}=50(g)\\ b)\\ n_{Fe(OH)_2}=n_{FeCl_2}=0,1(mol)\\ 4Fe(OH)_2+O_2\xrightarrow{t^o}2Fe_2O_3+4H_2O\\ n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe(OH)_2}=0,05(mol)\\ m=m_{Fe_2O_3}=0,05.160=8(g) \end{array}\)