Đặt y = x + 4 , khi đó phương trình trở thành :

( y - 1 ) ⁴ + ( y + 1 )⁴ = 16

\(\Leftrightarrow\)\(y^4-4y^3+6y^2-14y+1+y^4+4y^3+6y^2+14y+1=16\)

\(\Leftrightarrow2y^4+12y^2+2=16\)

\(\Leftrightarrow y^4+6y^2+1=8\)

\(\Leftrightarrow y^4+6y^2-7=0\)

\(\Leftrightarrow\left(y^2-1\right)\left(y^2+7\right)=0\)

Vì \(y^2+7\ge7\)với mọi y

Nên \(\Rightarrow y^2-1=0\)

\(\Leftrightarrow y^2=1\Rightarrow y=+-1\)

Do đó : x = -3 hoặc x = -5

Vậy S = { -3 ; -5 }

a) Ta có : d_{A/H2 = 8 \(\Leftrightarrow\)}\(\frac{M_A}{2}=8\Rightarrow M_A=8\cdot2=16\)

\(m_C=\frac{75\cdot16}{100}=12\left(g\right)\)

\(\Rightarrow n_C=\frac{m}{M}=\frac{12}{12}=1\left(mol\right)\)

\(\%H=100\%-\%C=100\%-75\%=25\%\)

\(m_H=\frac{25\cdot16}{100}=4\left(g\right)\)

\(\Rightarrow n_H=\frac{m}{M}=\frac{4}{1}=4\left(mol\right)\)

Vậy CTHH của A là : CH₄

b) \(n_{CH4}=\frac{m}{M}=\frac{4,8}{16}=0,3\left(mol\right)\)

PTPU : CH₄ + 2O₂ \(\rightarrow\) 2H₂O + CO₂

Theo phương trình , ta có : n_O2= n_{CH4\(\cdot\frac{2}{1}\)= 0,3 * 2/1 = 0,6 ( mol )}

_{\(\Rightarrow\)} V_O2 = n * 22,4 = 0,6 * 22,4 = 13,44 ( lít )

Vậy cần tối thiểu 13,44 lít khí O₂ để đốt cháy 4,8 g CH₄

a) Ta có : 2x + 6 = 2( x+ 3 )

x² + 3x = x ( x + 3 )

Nên \(\Rightarrow\)MTC = 2x ( x + 3 )

\(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}=\frac{\left(x+1\right)x+2\left(2x+3\right)}{2x\left(x+3\right)}=\frac{x^2+x+4x+6}{2x\left(x+3\right)}=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)

b) Ta có : 2x + 6 = 2( x + 3 )

2x² + 6x = 2x( x + 3 )

Nên : MTC = 2x ( x + 3 )

Do đó : \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{3x-x+6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)

Ta có : \(\frac{17}{3}=5+\frac{2}{3}=5+\frac{1}{\frac{3}{2}}=5+\frac{1}{1+\frac{1}{2}}\)

Nên suy ra : m = 5 ; n = 1 ; p = 2