\(R1//R2\)

\(\Rightarrow\left\{{}\begin{matrix}Rtd=\dfrac{R1R2}{R1+R2}=8\Omega\\U2=U1=Ia1.R1=12V\Rightarrow Ia2=\dfrac{U2}{R2}=\dfrac{12}{40}=0,3A\\Ia=Ia1+Ia2=0,3+1,2=1,5A\\Uv=U2=12V\end{matrix}\right.\)

\(n_{H_2}=\dfrac{3,36}{22,4}=0,15(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ Al_2O_3+6HCl\to 2AlCl_3+3H_2O\\ \Rightarrow n_{Al}=\dfrac{2}{3}n_{H_2}=0,1(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ \Rightarrow \%_{Al}=\dfrac{2,7}{5}.100\%=54\%\\ \Rightarrow \%_{Al_2O_3}=100\%-54\%=46\%\)

nH2=3,36/22,4=0,15

Đặt x,y lần lượt là số mol của al và al2o3

 2Al+6Hcl----->2AlCl3+3H2

 x                                   3/2x

Al2O3+6Hcl---->2AlCl3+3H2O

y                        

=>27x+102y=5(1)

3/2x=0,15(2)

Từ (1),(2)=>x=0,1,y=0,023

%Al=0,1*27/5*100%=54%

%Al2O3=100%-54%=46%

nH2=3,36/22,4=0,15

Đặt x,y lần lượt là số mol của Al,Al2O3

 2Al+6Hcl----->2AlCl3+3H2

 x                                   3/2x

Al2O3+6Hcl---->2AlCl3+3H2O

y                        

=>27x+102y=5(1)

3/2x=0,15(2)

Từ (1),(2)=>x=0,1,y=0,023

%Al=0,1*27/5*100%=54%

%Al2O3=100%-54%=46%

a: Ta có: \(\sqrt{x^2-4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\left|x-2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-2\\2x-3=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{3}\end{matrix}\right.\)

c: Ta có: \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\)

\(\Leftrightarrow\left|2x-1\right|=\left|x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x-3\\2x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\overrightarrow{AB}=\left(4;-7\right)\\\overrightarrow{BC}=\left(4;8\right)\\\overrightarrow{AC}=\left(8;1\right)\end{matrix}\right.\) \(\Rightarrow\overrightarrow{BC}+\overrightarrow{AC}=\left(12,9\right)\)

\(\Rightarrow\overrightarrow{AB}.\left(\overrightarrow{BC}+\overrightarrow{AC}\right)=4.12-7.9=...\)

b. Gọi \(H\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AH}=\left(x+3;y-5\right)\\\overrightarrow{BH}=\left(x-1;y+2\right)\\\end{matrix}\right.\)

Do \(AH\perp BC\Rightarrow\overrightarrow{AH}.\overrightarrow{BC}=0\)

\(\Rightarrow4\left(x+3\right)+8\left(y-5\right)=0\)

\(\Rightarrow x+2y=7\) (1)

Do H thuộc BC \(\Rightarrow\dfrac{x-1}{4}=\dfrac{y+2}{8}\Rightarrow2x-y=4\Rightarrow y=2x-4\)

Thế vào (1) \(\Rightarrow x+2\left(2x-4\right)=7\Rightarrow x=3\Rightarrow y=2\)

\(\Rightarrow H\left(3;2\right)\)