Giải pt

\(x^2-x-2=\sqrt{3-x}+\sqrt{x}\)

\(\Rightarrow\left(2x+1\right)\left(y-3\right)=15=1\cdot15=3\cdot5\)

Với \(x\in N\Rightarrow2x+1\ge1;2x+1\) lẻ

\(\left\{{}\begin{matrix}2x+1=1\\y-3=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=18\end{matrix}\right.\\ \left\{{}\begin{matrix}2x+1=15\\y-3=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=7\\y=4\end{matrix}\right.\\ \left\{{}\begin{matrix}2x+1=3\\y-3=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=8\end{matrix}\right.\\ \left\{{}\begin{matrix}2x+1=5\\y-3=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3\\y=6\end{matrix}\right.\)

Vậy \(x\in\left\{0;1;3;7\right\}\)

\(n_{O_2}=\dfrac{2,479}{24,79}=0,1(mol)\\ \Rightarrow A\)

\(a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{H_2}=\dfrac{24,79}{24,79}=1(mol)\\ \Rightarrow m_{H_2}=1.2=2(g)\\ \text {Bảo toàn KL: }m_{HCl}+m_{Zn}=m_{ZnCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}=136+24-2=140,5(g)\\ c,PTHH:H_2+CO_2\xrightarrow{t^o}CO+H_2O\\ H_2+Cl_2\xrightarrow{t^o}2HCl\)

Vì \(\dfrac{n_{H_2}}{1}>\dfrac{n_{CO_2}}{1};\dfrac{n_{H_2}}{1}>\dfrac{n_{Cl_2}}{1}\) nên sau phản ứng \(H_2\) dư

\(\Rightarrow \begin{cases} n_{CO}=0,5(mol\\ n_{HCl}=2n_{Cl_2}=0,7(mol) \end{cases}\\ \Rightarrow m_{hh}=m_{CO}+m{HCl}=0,5.28+0,7.36,5=39,55(g)\\ V_{hh}=V_{CO}+V_{HCl}=0,5.22,4+0,7.22,4=26,88(l)\)

Thầy ơi sao bên này là (2-mn) qua bên kia lại là \(\left[8-\left(m+n\right)^2\right]\) , dưới mẫu là (m+1)(n+1) qua bên này là \(\text{(m+n+2)}^2\)