\(a,n_{Al}=\dfrac{6,75}{27}=0,25(mol)\\ PTHH:2Al+6HCl\to 2AlCl_3+3H_2\\ \Rightarrow n_{AlCl_3}=n_{Al}=0,25(mol)\\ \Rightarrow m_{AlCl_3}=0,25.133,5=33,375(g)\\ b,n_{H_2}=1,5.n_{Al}=0,375(mol)\\ \Rightarrow V_{H_2}=0,375.22,4=8,4(l)\\ \Rightarrow V_{H_2(tt)}=8,4.90\%=7,56(l)\)

\(=x^2\left(x-1\right)-4\left(x-1\right)^2=\left(x-1\right)\left[x^2-4\left(x-1\right)\right]\\ =\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)

\(=4\cdot3125-2^5:2^4=12500-2=12498\)

\(a,A=\dfrac{2^2-9}{3\left(2+5\right)}=\dfrac{-5}{21}\\ b,B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}\\ B=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\\ c,P=AB=\dfrac{\left(x-3\right)\left(x+3\right)}{3\left(x+5\right)}\cdot\dfrac{3}{x+3}=\dfrac{x-3}{x+5}\\ P=\dfrac{x+5-8}{x+5}=1-\dfrac{8}{x+5}\in Z\\ \Leftrightarrow x+5\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow x\in\left\{-13;-9;-7;-6;-4;-1\right\}\left(x\ne\pm3\right)\)

Gọi 2 số tự nhiên lt là \(a,a+1(a\in \mathbb{N^*})\)

Ta có \(a\left(a+1\right)=600\)

Ta lại có \(600=2^3\cdot3\cdot5^2=\left(2^3\cdot3\right)\cdot5^2=24\cdot25\)

\(\Rightarrow\left\{{}\begin{matrix}a=24\\a+1=25\end{matrix}\right.\Rightarrow a=24\)

Vậy  2 số đó là 24,25

\(13,44(dm^3)=13,44(l)\\ a,n_{SO_2}=\dfrac{13,44}{22,4}=0,6(mol)\\ b,m_{SO_2}=0,6.64=38,4(g)\\ c,n_{O}=2n_{SO_2}=1,2(mol)\\ \text{Số nguyên tử oxi: }1,2.6.10^{23}=7,2.10^{23}\\ d,\text{Số phân tử }H_2=5.\text{Số phân tử }SO_2\\ \Rightarrow n_{H_2}=5n_{SO_2}=3(mol)\\ \Rightarrow m_{H_2}=3.2=6(g)\)

PT hoành độ giao điểm: \(2mx+m-1=x+1\)

2 đt Cắt trên Ox \(\Leftrightarrow x=0\Leftrightarrow2m\cdot0+m-1=0+1=1\Leftrightarrow m=2\)

2 đt Cắt trên Oy \(\Leftrightarrow\left\{{}\begin{matrix}2mx+m-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2m+m-1=0\\x=-1\end{matrix}\right.\Leftrightarrow m=-1\) 

\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a+b}{8+3}=\dfrac{33}{11}=3\\ \Rightarrow\left\{{}\begin{matrix}a=24\\b=9\end{matrix}\right.\)

Vì x,y tỉ lệ nghịch nên \(x_1y_1=x_2y_2\)

\(\Rightarrow\dfrac{y_1}{x_2}=\dfrac{y_2}{x_1}=\dfrac{y_1}{2}=\dfrac{y_2}{5}=\dfrac{y_2+y_1}{2+5}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}y_1=2\cdot3=6\\y_2=3\cdot5=15\end{matrix}\right.\)

\(S_{ABC}=\dfrac{1}{2}\times AF\times BC=\dfrac{1}{2}\times\left(2+3\right)\times\left(2+4\right)=\dfrac{1}{2}\times5\times6=15\left(cm^2\right)\\ S_{BEC}=\dfrac{1}{2}\times EF\times BC=\dfrac{1}{2}\times3\times6=9\left(cm^2\right)\\ \Rightarrow S_{ABEC}=S_{ABC}-S_{BEC}=15-9=6\left(cm^2\right)\)