\(\dfrac{ay-bx}{c}=\dfrac{cx-az}{b}=\dfrac{bz-cy}{a}\\ \Rightarrow\dfrac{acy-bcx}{c^2}=\dfrac{bcx-abz}{b^2}=\dfrac{abz-acy}{a^2}=\dfrac{acy-bcx+bxc-abz+abz-acy}{a^2+b^2+c^2}=0\\ \Rightarrow\left\{{}\begin{matrix}ay-bx=0\\cx-az=0\\bz-cy=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}ay=bx\\cx=az\\bz=cy\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{a}=\dfrac{y}{b}\\\dfrac{x}{a}=\dfrac{z}{c}\\\dfrac{y}{b}=\dfrac{z}{c}\end{matrix}\right.\\ \Rightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

\(a,=5\sqrt{2}-3\sqrt{2}+\sqrt{2}=3\sqrt{2}\\ b,=\dfrac{\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}=\dfrac{2\sqrt{3}}{3-2}=2\sqrt{3}\)

\(ĐK:x^2-5x-6\ge0\\ PT\Leftrightarrow\sqrt{x^2-5x-6}+2\left(x^2-5x-6\right)=0\\ \Leftrightarrow\sqrt{x^2-5x-6}\left(1+2\sqrt{x^2-5x-6}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-5x-6=0\left(tmĐK\right)\\2\sqrt{x^2-5x-6}=-1\left(vn\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

Đổi \(18m^2=180000cm^2\)

Số gạch cần dùng là \(180000:\left(30\times30\right)=200\) viên gạch

\(a,\text{Sơ đồ p/ứ: }Fe+HCl\to FeCl_2+H_2\\ b,PTHH:Fe+2HCl\to FeCl_2+H_2\\ c,\text{Bảo toàn KL: }m_{Fe}+m_{HCl}=m_{FeCl_2}+m_{H_2}\\ \Rightarrow m_{HCl}+56=150+8=158\\ \Rightarrow m_{HCl}=102(g)\)

Đặt KL trung bình là R \(\rightarrow \) R hóa trị II

\(R+2HCl\to RCl_2+H_2\\ \Rightarrow n_R=n_{RCl_2}\\ \Rightarrow \dfrac{31,4}{M_R}=\dfrac{52,7}{M_R+71}\\ \Rightarrow 21,3M_R=2229,4\\ \Rightarrow M_R=104,67\)

Vậy 2 KL đó là Sr và Ba

\(\Rightarrow4x^3=515-15=500\\ \Rightarrow x^3=500:4=125=5^3\\ \Rightarrow x=5\)

\(\widehat{F}=90^0-\widehat{E}=30^0\)

\(DE=\tan F\cdot DF=\tan30^0\cdot10=\dfrac{\sqrt{3}}{3}\cdot10=\dfrac{10\sqrt{3}}{3}\left(cm\right)\\ EF=\dfrac{DE}{\sin F}=\dfrac{\dfrac{10\sqrt{3}}{3}}{\sin30^0}=\dfrac{20\sqrt{3}}{3}\left(cm\right)\)

\(=\dfrac{x+3+4x-12-2x}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

\(n_{Fe}=\dfrac{6,8}{56}=\dfrac{17}{140}(mol)\\ PTHH:3Fe+2O_2\xrightarrow{t^o}Fe_3O_4\\ a,n_{O_2}=\dfrac{2}{3}n_{Fe}=\dfrac{17}{210}(mol)\\ \Rightarrow V_{O_2}=\dfrac{17}{210}.22,4=1,81(g)\\ b,n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=\dfrac{17}{420}(mol)\\ \Rightarrow m_{Fe_3O_4}=\dfrac{17}{420}.232=9,39(g)\)