Ta cho x<y mà x,y là số nguyên liên tiếp

=> x=y-1

\(y^2-x^2>20\)

\(\left(y+x\right)\left(y-x\right)>20\)

\(\left(y+y-1\right)\left(y-\left(y-1\right)\right)>20\)

\(2y-1>20\)

\(2y>21\)

\(y>10,5\)

Mà y là số nguyên nên để đạt GTNN thì y=11

=> x=11-1=10

GTNN: \(y^2+x^2=11^2+10^2=221\)

2H₂+O₂==>2H₂O

\(n_{H_2}=\frac{8}{2}=4mol\)

\(n_{O_2}=\frac{48}{32}=1,5mol\)

\(\frac{4}{2}>\frac{1,5}{1}\)=> H₂ dư

\(n_{H_2O}=2.n_{O_2}=2.1,5=3mol\)

\(m_{H_2O}=3.18=54g\)

=x⁴-x²+4x²-4

=x²(x²-1)+4(x²-1)

=(x²-1)(x²+4)

=(x-1)(x+1)(x²+4)

***

Đúng cho mk nha

a) PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b) \(n_{Al}=\frac{8,1}{27}=0,3mol\)

\(n_{HCl}=\frac{21,5}{36,5}=\frac{43}{73}\left(mol\right)\)

\(\frac{0,3}{2}>\frac{\frac{47}{73}}{6}\)=> Al dư

\(n_{Al\left(dư\right)}=0,3-\left(\frac{\frac{43}{73}.2}{6}\right)=\frac{227}{2190}mol\)

\(m_{Al\left(dư\right)}=\frac{227}{2190}.27=\frac{2043}{730}g\)

\(n_{AlCl_3}=\frac{2}{6}.n_{HCl}=\frac{2}{6}.\left(\frac{43}{73}\right)=\frac{43}{219}mol\)

\(m_{AlCl_3}=\frac{43}{219}.133,5=\frac{3827}{146}g\)