\(a,ĐK:x\ne\pm3\\ b,P=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}=\dfrac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{4}{x-3}\\ c,P=4\Leftrightarrow\dfrac{4}{x-3}=4\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)

\(x^2+y^2+z^2=xy+yz+zx\\ \Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\\ \Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y-z=0\\z-x=0\end{matrix}\right.\Leftrightarrow x=y=z\\ \text{Mà }x+y+z=-3\Leftrightarrow x=y=z=-1\\ \Leftrightarrow B=1-1+1=1\)

Vì Parabol có đỉnh S(1;5) nên \(-\dfrac{b}{2a}=-\dfrac{2}{2a}=-\dfrac{1}{a}=1\Leftrightarrow a=-1\)

\(\left(P\right):y=-x^2+2x+c\) đi qua \(S\left(1;5\right)\)

\(\Leftrightarrow-1+2+c=5\Leftrightarrow c=4\)

Vậy \(\left(P\right):y=-x^2+2x+4\)

Áp dụng định luật BTKL, ta có:

\(m_S+m_{O_2}=m_{SO_2}\\ \Rightarrow m_{O_2}=16-8=8(g)\\ \Rightarrow n_{O_2}=\dfrac{8}{32}=0,25(mol)\\ \Rightarrow V_{O_2}=0,25.22,4=5,6(l)\)

\(1,=\left(x^2-2x+6x-12+2\right):\left(x-2\right)\\ =\left[x\left(x-2\right)+6\left(x-2\right)+2\right]:\left(x-2\right)\\ =x+6\left(\text{dư }2\right)\\ 2,=\left(x^3-x^2+5x^2-5x+6x-6\right):\left(x-1\right)\\ =\left[x^2\left(x-1\right)+5x\left(x-1\right)+6\left(x-1\right)\right]:\left(x-1\right)\\ =x^2+5x+6\)

Tổng tiền ban đầu là \(470\times5000=2350000\left(đồng\right)\)

Số tiền được bớt là \(2350000\times4\%=94000\left(đồng\right)\)

Câu 2:

\(\begin{array}{|c|c|c|}\hline \text{}&HCl&Na_2SO_4&NaOH\\\hline \text{Quỳ tím}&\text{đỏ}&x&\text{xanh}\\\hline\end{array}\)

Câu 3:

\(a,PTHH:Fe+2HCl\to FeCl_2+H_2\\ n_{H_2}=\dfrac{6,72}{22,4}=0,3(mol)\\ b,n_{Fe}=n_{H_2}=0,3(mol)\\ \Rightarrow m_{Fe}=0,3.56=16,8(g)\\ \Rightarrow \%m_{Fe}=\dfrac{16,8}{30}.100\%=56\%\\ \Rightarrow \%m_{Cu}=100\%-56\%=44\%\)

Vì \(x\neq y\) nên Fe hóa trị III do SO₄ hóa trị II cố định

Theo quy tắc hóa trị ta có:

\(x.III=y.II\Rightarrow \dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow CTHH:Fe_2(SO_4)_3\\ PTHH:3Fe(OH)_3+3H_2SO_4\to Fe_2(SO_4)_3+6H_2O\)

Oxit Bazo: \(CuO,Fe_2O_3\)

Oxit Axit: \(SO_2,P_2O_5\)

Oxit trung tính: \(CO,NO\)

Oxit lưỡng tính: \(Al_2O_3\)

Điều chế SO₂:

\(Na_2SO_3+H_2SO_4\to Na_2SO_4+H_2O+SO_2\uparrow\)

Sản xuất CaO:

\(CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow\)