\(n_{CO_2}=\dfrac{6,6}{44}=0,15(mol)\\ V_{CO_2}=0,15.22,4=6,72(l)\\ n_{SO_2}=\dfrac{6,4}{64}=0,1(mol)\\ V_{SO_2}=0,1.22,4=2,24(l)\\ n_{Cl_2}=\dfrac{14,2}{35,5}=0,4(mol)\\ V_{Cl_2}=0,4.22,4=8,96(l)\)

Trừ 2 vế của HPT

\(\Leftrightarrow x^2-xy+y^2-x+y-xy=0\\ \Leftrightarrow x^2+y^2-x+y-2xy=0\\ \Leftrightarrow\left(x-y\right)^2-\left(x-y\right)=0\\ \Leftrightarrow\left(x-y\right)\left(x-y-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=y\\x=y+1\end{matrix}\right.\)

Với \(x=y\Leftrightarrow x-x+x^2=7\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\Rightarrow y=\sqrt{7}\\x=-\sqrt{7}\Rightarrow y=-\sqrt{7}\end{matrix}\right.\)

Với \(x=y+1\Leftrightarrow y+1-y+y\left(y+1\right)=7\)

\(\Leftrightarrow y^2+y-6=0\\ \Leftrightarrow\left[{}\begin{matrix}y=2\Rightarrow x=3\\y=-3\Rightarrow x=-2\end{matrix}\right.\)

Vậy ...

\(S=5,99.0,1=0,599\left(km\right)\)

\(n_{CO_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ \text{Bảo toàn nguyên tố C,H}:\\ n_{C(A)}=n_{CO_2}=0,3(mol)\\ \Rightarrow m_{C(A)}=0,3.12=3,6(g)\\ n_{H(A)}=2n_{H_2O}=0,8(mol)\\ \Rightarrow m_{H(A)}=0,8.1=0,8(g)\\ \Rightarrow m_{O(A)}=9-3,6-0,8=4,6(g)\\\)

\(a,=\dfrac{11}{12}\cdot\dfrac{16}{33}\cdot\dfrac{3}{5}=\dfrac{4}{15}\\ b,=\dfrac{2^2\cdot5^2\cdot5^3}{2^3\cdot5^4}=\dfrac{5}{2}\\ c,=1+\dfrac{7}{3}-\dfrac{5}{3}=\dfrac{5}{3}\\ d,=5^3\cdot\dfrac{2^4}{5^4}+\dfrac{14}{5}=\dfrac{2^4}{5}+\dfrac{14}{5}=\dfrac{30}{5}=6\)

\(a,m^2-2m+3=\left(m-1\right)^2+2\ge2>0\) nên \(\left(d\right)\) tạo với \(Ox\) một góc nhọn

\(b,\Leftrightarrow\left\{{}\begin{matrix}m^2-2m+3=3\\2k-1\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\\k\ne1\end{matrix}\right.\)

\(\Leftrightarrow4\left(n+2\right)+2⋮n+2\\ \Leftrightarrow n+2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow n=0\left(n\in N\right)\)

\(=x-2008-\sqrt{x-2008}+\dfrac{1}{4}+\dfrac{8031}{4}\\ =\left(\sqrt{x-2008}-\dfrac{1}{2}\right)^2+\dfrac{8031}{4}\ge\dfrac{8031}{4}\)

Dấu \("="\Leftrightarrow\sqrt{x-2008}=\dfrac{1}{2}\Leftrightarrow x-2008=\dfrac{1}{4}\Leftrightarrow x=\dfrac{8033}{4}\)

\(i,\Leftrightarrow\left\{{}\begin{matrix}x+2y=5\sqrt{5}\\2\sqrt{5}x+2y=10+4\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(2\sqrt{5}-1\right)x=10-\sqrt{5}\\x+2y=5\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{10-\sqrt{5}}{2\sqrt{5}-1}=\dfrac{\sqrt{5}\left(2\sqrt{5}-1\right)}{2\sqrt{5}-1}=\sqrt{5}\\\sqrt{5}+2y=5\sqrt{5}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{5}\\y=2\sqrt{5}\end{matrix}\right.\)