c: \(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{1}{x-1}\)

\(\dfrac{x+y}{2\left(x-y\right)}+\dfrac{2y^2}{y-x}=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{2y^2}{x-y}=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{4y^2}{2\left(x-y\right)}=\dfrac{x+y-4y^2}{x-y}\)

x⁴+x²+1

=(x²)²+2x²+1-2x²+x²

=(x²+1)²-2x²+x² 

= (x² + 1)² − x² 

= (x² + x+ 1 )(x² − x+ 1 )

\(x^4+x^2+1\)
\(=\left[\left(x^2\right)^2+2.x^2.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]-\left(\frac{1}{2}\right)^2+1\)
\(=\left(x^2+\frac{1}{2}\right)^2-\frac{1}{4}+\frac{4}{4}\)
\(=\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}\)

\(=\dfrac{x^2+2+5x-10}{x^2-4}=\dfrac{x^2+5x-8}{\left(x-2\right)\left(x+2\right)}\)

Đặt CTHH: C_xH_yO_zN_t

\(n_C=n_{CO_2}=\frac{11}{44}=0,25mol\)

\(n_H=2n_{H_2O}=2.\left(\frac{6,3}{18}\right)=0,7mol\)

Tổng số mol N( trong hợp chất hữu cơ+trong không khí )= 2 lần số mol N₂=\(2.\left(\frac{34,72}{22,4}\right)=3,1mol\)

Tổng số mol O( trong hợp chất hữu cơ+trong không khí )= 2 lần số mol CO₂+số mol nước=\(2.0,25+0,35=0,85mol\)

Trong không khí: \(\left\{\begin{matrix}N_2=4a\left(mol\right)\\O_2=a\left(mol\right)\end{matrix}\right.\)=> Trong hợp chất hữu cơ thì \(\left\{\begin{matrix}n_{N\left(trongchathuuco\right)}=3,1-4a\\n_{O\left(trongchathuuco\right)}=0,85-a\end{matrix}\right.\)

\(m_{N\left(trongchathuuco\right)}+m_{O\left(trongchathuuco\right)}=m_{chathuuco}-m_{C\left(trongchathuuco\right)}-m_{H\left(trongchathuuco\right)}\)

\(14.\left(3,1-4a\right)+16.\left(0,85-a\right)=6,7-12.0,25-1.0,7\Rightarrow a=0,75\)

Theo hệ pt: Ta có: \(n_{N\left(trongchathuuco\right)}=3,1-4.0,75=0,1mol\)

\(n_{O\left(trongchathuuco\right)}=0,85-0,75=0,1\)

\(x:y:z:t=0,25:0,7:0,1:0,1=5:14:2:2\)=>CTHH: C₅H₁₄O₂N₂

\(\dfrac{x}{x-y}-\dfrac{1}{x-y}-\dfrac{1-y}{y-x}=\dfrac{x}{x-y}-\dfrac{1}{x-y}+\dfrac{y-1}{x-y}=\dfrac{x-1+y-1}{x-y}=\dfrac{x+y-2}{x-y}\)