\(a,CTTQ:K_x^{I}(SO_4)_y^{II}\\ \Rightarrow x.I=y.II\Rightarrow \dfrac{x}{y}=2\Rightarrow x=2;y=1\\ \Rightarrow K_2SO_4\\ b,CTTQ:Al_x^{III}(NO_3)_y^{I}\\ \Rightarrow x.III=y.I\Rightarrow \dfrac{x}{y}=\dfrac{1}{3}\Rightarrow x=1;y=3\\ \Rightarrow Al(NO_3)_3\\ c,CTTQ:Fe_x^{III}(OH)_y^I\\ \Rightarrow x.III=y.I\Rightarrow \dfrac{x}{y}=\dfrac{1}{3}\Rightarrow x=1;y=3\\ \Rightarrow Fe(OH)_3\\ \)

\(d,CTTQ:Ba_x^{II}(PO_4)_y^{III}\\ \Rightarrow x.II=y.III\Rightarrow \dfrac{x}{y}=\dfrac{3}{2}\Rightarrow x=3;y=2\\ \Rightarrow Ba_3(PO_4)_2\)

\(CuSO_4+2NaOH\to Cu(OH)_2+Na_2SO_4\\ Cu(OH)_2\xrightarrow{t^o}CuO+H_2O\\ \Rightarrow n_{CuO}=0,2(mol)\Rightarrow m=0,2.80=16(g)\)

\(2M+3Cl_2\Rightarrow 2MCl_3\\ \Rightarrow n_M=n_{MCl_3}\\ \Rightarrow \dfrac{10,8}{M_M}=\dfrac{53,4}{M_M+35,5.3}\\ \Rightarrow M_M=27\Rightarrow Al\)

Bs: \(x,y\in \mathbb{Z}\)

Ta có \(36-y^2=8\left(x-2021\right)^2\ge0\Leftrightarrow y^2\le36\)

Mà \(8\left(x-2021\right)^2\) và 36 chẵn nên y chẵn

Do đó \(y^2\in\left\{4;16;36\right\}\)

Với \(y^2=4\Leftrightarrow8\left(x-2021\right)^2=32\Leftrightarrow\left(x-2021\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x=2025\\x=2017\end{matrix}\right.\)

Với \(y^2=16\Leftrightarrow8\left(x-2021\right)^2=20\Leftrightarrow\left(x-2021\right)^2=\dfrac{5}{2}\left(loại\right)\)

Với \(y^2=36\Leftrightarrow8\left(x-2021\right)^2=0\Leftrightarrow x=2021\)

Vậy \(\left(x;y\right)=\left(2025;2\right);\left(2025;-2\right);\left(2017;2\right);\left(2017;-2\right);\left(2021;6\right);\left(2021;-6\right)\)

\(n_S=\dfrac{6,4}{32}=0,2(mol)\\ S+O_2\xrightarrow{t^o}SO_2\\ \Rightarrow n_{SO_2}=0,2(mol)\\ \Rightarrow V_{SO_2}=0,2.22,4=4,48(l) \)

\(a,2H_2+O_2\xrightarrow{t^o}2H_2O\\ b,n_{H_2}=\dfrac{4,48}{22,4}=0,2(mol)\\ \Rightarrow n_{H_2O}=0,2(mol)\\ \Rightarrow m_{H_2O}=0,2.18=3,6(g)\\ c,n_{O_2}=0,1(mol)\\ \Rightarrow V_{O_2}=0,1.22,4=2,24(l)\)

\(a,B=\dfrac{-2}{2\cdot4+1}=-\dfrac{2}{9}\\ b,A=\dfrac{x^2-3x+2x+6-x^2-3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{-1}{x+3}\\ c,P=2A:B=\dfrac{-2}{x+3}\cdot\dfrac{2x+1}{-2}=\dfrac{2x+1}{x+3}=2-\dfrac{5}{x+3}\in Z\\ \Leftrightarrow x+3\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\\ \Leftrightarrow x\in\left\{-8;-4;-2;2\right\}\left(tm\right)\)

\(\left\{{}\begin{matrix}AB=AC\\BD=DC\\AD\text{ chung}\end{matrix}\right.\Rightarrow\Delta ADB=\Delta ADC\left(c.c.c\right)\\ \Rightarrow\widehat{BAD}=\widehat{CAD}\)

Vậy AD là p/g \(\widehat{BAC}\)

Gọi 3 phần là a,b,c

Áp dụng tc dtsbn:

\(3a=5b=7c\Rightarrow\dfrac{3a}{105}=\dfrac{5b}{105}=\dfrac{7c}{105}\\ \Rightarrow\dfrac{a}{35}=\dfrac{b}{21}=\dfrac{c}{15}=\dfrac{a+b+c}{35+21+15}=\dfrac{284}{71}=4\\ \Rightarrow\left\{{}\begin{matrix}a=140\\b=84\\c=60\end{matrix}\right.\)

Vậy ...