\(\dfrac{x+y-2017z}{z}=\dfrac{y+z-2017x}{x}=\dfrac{z+x-2017y}{y}\)

<=> \(\dfrac{x+y}{z}-2017=\dfrac{z+y}{x}-2017=\dfrac{z+x}{y}-2017\)

<=> \(\dfrac{x+y}{z}=\dfrac{z+y}{x}=\dfrac{z+x}{y}\)

đặt x+y+z = t 

=> \(\dfrac{t-z}{z}=\dfrac{t-x}{x}=\dfrac{t-y}{y}< =>\dfrac{t}{z}-1=\dfrac{t}{x}-1=\dfrac{t}{y}-1\) \(< =>\dfrac{t}{z}=\dfrac{t}{y}=\dfrac{t}{x}\)

=> x=y=z 

ta lại có 

\(P=\left(1+\dfrac{y}{x}\right)\left(1+\dfrac{x}{z}\right)\left(1+\dfrac{z}{y}\right)\)

vì x=y=z  => P = \(\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)

\(a.CaCO_3-^{t^o}>CaO+CO_2\)

\(b.C+O_2-^{t^o}>CO_2\)

\(c.Zn+2HCl->ZnCl_2+H_2\)

\(d.2H_2+O_2-^{t^o}>2H_2O\)

P+E+N=34 => 2P+N=34

N =12 

=> P=11 => x là Natri (Na)

X.1=II . 2 

=> x =4 => hóa trị của Mn là IV

sửa đề Al2O3

X.2=II . 3 => x=3 => hóa trị của Al là III

B Mg; Al ; Zn 

\(Mg+H_2SO_4->MgSO_4+H_2\)

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

\(Zn+H_2SO_4->ZnSO_4+H_2\)

\(1.2Fe+O_2-^{t^o}>2FeO\)

\(2.FeO+2HCl-^{t^o}>FeCl_2+H_2O\)

\(3.FeCl_2+Ba\left(OH\right)_2->Fe\left(OH\right)_2+BaCl_2\)

\(4.Fe\left(OH\right)_2-^{t^o}>FeO+H_2O\)

\(5.FeO+H_2->Fe+H_2O\)

a) Fe2O3 

Fe2O3 + 6HCl -> 2FeCl3 + 3H2O

b) SO3 , CO2 

Ca(OH)2 + CO2 -> CaCO3 + H2O

Ca(OH)2 + SO3 -> CaSO4 + H2O

c) Cu(OH)2 

Cu(OH)2 -> CuO +H2O

d) Fe2O3 , CuO 

\(CuO+H2O-^{t^o}>Cu\left(OH\right)2\)

\(Fe2O3+3H2O-^{t^o}>2Fe\left(OH\right)3\)

cho 2 số thực dương a,b sao cho \(\sqrt{a+1}+\sqrt{b+1}=4\)

tìm Min P = \(a^4+b^4\)