Đặt \(y=\frac{x^2+x+1}{x^2+2x+1}\)

\(x^2y+2xy+y=x^2+x+1\)

\(\left(1-y\right)x^2+\left(1-2y\right)x+\left(1-y\right)=0\)

\(\Delta=\left(1-2y\right)^2-4\left(1-y\right)^2\)

\(\Delta=4y-3\)

\(4y-3\ge0\)

Để đạt GTNN thì \(y=\frac{3}{4}\)

\(Min_Q=\frac{3}{4}\) tại x=1

10l=0,01m³

\(D_{cát}=\frac{m}{V}=\frac{15}{0,01}=1500\left(\frac{kg}{m^3}\right)\)

\(d=10D=10.1500=15000\left(\frac{N}{m^3}\right)\)

\(P=d.V=15000.1=15000N\)

4P+5O₂=t^o=>2P₂O₅

\(V_{O_2}=\frac{3,36}{5}=0,672l\)

\(n_{O_2}=\frac{0,672}{22,4}=0,03mol\)

\(m_{O_2}=0,03.32=0,96g\)

\(n_{P_2O_5}=\frac{2}{5}.n_{O_2}=\frac{2}{5}.0,03=0,012mol\)

\(m_{P_2O_5}=0,012.142=1,704g\)

4P+5O₂=t^o=>2P₂O₅

\(n_P=\frac{6,2}{31}=0,2\left(mol\right);n_{O_2}=\frac{2,24}{22,4}=0,1mol\)

\(\frac{0,2}{4}>\frac{0,1}{5}\)=>P dư

a) \(n_{P\left(dư\right)}=0,2-\left(\frac{0,1.4}{5}\right)=0,12mol\)

\(m_{P\left(dư\right)}=0,12.31=3,72g\)

b) \(n_{P_2O_5}=\frac{2}{5}.n_{O_2}=\frac{2}{5}.0,1=0,04mol\)

\(m_{P_2O_5}=0,04.142=5,68g\)

c) 2KMnO₄-np->K₂MnO₄+MnO₂+O2

\(n_{KMnO_4}=2.n_{O_2}=2.0,1=0,2mol\)

\(m_{KMnO_4}=0,2.158=31,6g\)

2M + 2nHCl->2MCln + n H2
0,4/n<-0,4-> 0,4/n->0,2
m dd =0,4M/n +200-0,4=0,4M/n +199,6 g
m muối =(M+35,5n).0,4/n
C%=11,96%
-> M=27,47n ???????????
-> Mn

\(x^3+y^3=108\)

\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)=108\)

\(\Rightarrow x^2-xy+y^2=18\left(1\right)\)

\(x+y=6\Rightarrow\left(x+y\right)^2=36\Leftrightarrow x^2+2xy+y^2=36\left(2\right)\)

Lấy (2)-(1) ta được 3xy=18=>xy=6