\(\begin{array}{|c|c|c|} \hline &SO_2&PbO&H_2O&BaO&N_2O&CuO&ZnO&SO_3&CaO\\ \hline K_2O&\rm x&&\rm x&&&&&&\\ \hline H_2SO_4&&\rm x&&\rm x&&\rm x&\rm x&&\rm x\\ \hline NaOH&\rm x&\rm x&&&&&\rm x&\rm x\\ \hline\end{array}\)

$K_2O+SO_2\to K_2SO_3$
$K_2O+H_2O\to 2KOH$
$PbO+H_2SO_4\to PbSO_4+H_2O$

$BaO+H_2SO_4\to BaSO_4\downarrow+H_2O$

$CuO+H_2SO_4\to CuSO_4+H_2O$

$ZnO+H_2SO_4\to ZnSO_4+H_2O$

$CaO+H_2SO_4\to CaSO_4+H_2O$

$SO_2+2NaOH\to Na_2SO_3+H_2O$

$PbO+2NaOH\to Na_2PbO_2+H_2O$

$ZnO+2NaOH\to Na_2ZnO_2+H_2O$

$SO_3+NaOH\to NaHSO_4$

Que đóm bùng cháy do xuất hiện khí oxi sau phản ứng $(O_2)$

$2KMnO_4\xrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow$

$n_{Na}=0,1(mol);n_{Zn}=0,1(mol)$

$2Na+S\xrightarrow{t^o}Na_2S$

$Zn+S\xrightarrow{t^o}ZnS$

$Na_2S+2HCl\to 2NaCl+H_2S$

$ZnS+2HCl\to ZnCl_2+H_2S$

Theo PT: $\sum n_{H_2S}=\frac{1}{2}n_{Na}+n_{Zn}=0,15(mol)$

$H_2S+Cu(NO_3)_2\to CuS\downarrow+2HNO_3$

Theo PT: $n_{CuS}=n_{H_2S}=0,15(mol)$

$\to a=0,15.96=14,4(g)$

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$

\(\dfrac{2000}{2001}\cdot\dfrac{2002}{2003}\cdot\dfrac{2001}{2002}\cdot\dfrac{2003}{2004}\cdot\dfrac{2006}{2000}=\dfrac{2006}{2004}=\dfrac{1003}{1002}\)

\(x^2+3x+5=xy+2y\\ \Leftrightarrow x^2+3x-xy-2y+5=0\\ \Leftrightarrow x\left(x+2\right)-y\left(x+2\right)+\left(x+2\right)+3=0\\ \Leftrightarrow\left(x+2\right)\left(x-y+1\right)=-3=\left(-1\right)\cdot3=\left(-3\right)\cdot1\)

\(TH_1:\left\{{}\begin{matrix}x+2=-3\\x-y+1=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-5\end{matrix}\right.\to\left(-5;-5\right)\\ TH_2:\left\{{}\begin{matrix}x+2=3\\x-y+1=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\to\left(1;3\right)\\ TH_3:\left\{{}\begin{matrix}x+2=1\\x-y+1=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=3\end{matrix}\right.\to\left(-1;3\right)\\ TH_4:\left\{{}\begin{matrix}x+2=-1\\x-y+1=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=-5\end{matrix}\right.\to\left(-3;-5\right)\)

Vậy \(\left(x;y\right)=\left(-5;-5\right);\left(1;3\right);\left(-1;3\right);\left(-3;-5\right)\)

\(a,\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow\dfrac{a^2}{c^2}=\dfrac{c^2}{b^2}=\dfrac{a^2+c^2}{b^2+c^2}\left(1\right)\)

Mà \(\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow ab=c^2\Leftrightarrow\dfrac{a}{b}=\dfrac{c^2}{b^2}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\tođpcm\)

\(b,\dfrac{a}{c}=\dfrac{c}{b}\Leftrightarrow ab=c^2\)

\(\Leftrightarrow\dfrac{b^2-a^2}{a^2+c^2}=\dfrac{\left(b-a\right)\left(b+a\right)}{a^2+ab}=\dfrac{\left(b-a\right)\left(b+a\right)}{a\left(a+b\right)}=\dfrac{b-a}{a}\left(đpcm\right)\)

\(\left\{{}\begin{matrix}A+B=1639\left(1\right)\\B+C=1528\left(2\right)\\C+A=1115\left(3\right)\end{matrix}\right.\\ \Rightarrow\left(1\right)+\left(2\right)+\left(3\right)=2\left(A+B+C\right)=4282\\ \Rightarrow A+B+C=2141\left(4\right)\\ \Rightarrow\left(4\right)-\left(2\right)=A+B+C-B-C=2141-1528\\ \Rightarrow A=613\)

Vậy $A=613$

\(\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right):\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(a>0;a\ne1\right)\\ =\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{a-1}{\sqrt{a}}\\ =\dfrac{4a\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\\ =\dfrac{4a^2}{\left(\sqrt{a}-1\right)^2\left(\sqrt{a}+1\right)^2}\)

\(a,m=2\\ PT\left(\text{*}\right)\Leftrightarrow x^2+4x-12=0\\ \Leftrightarrow x^2-2x+6x-12=0\\ \Leftrightarrow\left(x-2\right)\left(x+6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\\ b,\text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=-4\left(m-1\right)=4-4m\\x_1x_2=-12\end{matrix}\right.\\ \text{Ta có }4=\left(x_1+x_2-x_1x_2-8\right)^2\\ \Leftrightarrow\left(4-4m+12-8\right)^2=4\\ \Leftrightarrow\left(8-4m\right)^2=4\\ \Leftrightarrow\left(4-2m\right)^2=1\\ \Leftrightarrow\left[{}\begin{matrix}4-2m=1\\2m-4=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=\dfrac{3}{2}\\m=\dfrac{5}{2}\end{matrix}\right.\)