\(n_{HCl}=0,5.0,4=0,2\left(mol\right)\) ; \(n_{KOH}=\dfrac{80.2,8\%}{56}=0,04\left(mol\right)\)

\(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\) (1)

\(KOH+HCl\rightarrow KCl+H_2O\) (2)

Theo pt (2): \(\Rightarrow n_{HCl\left(dư\right)}=0,04\left(mol\right)\) \(\Rightarrow n_{HCl\left(p.ứ\right)}=0,2-0,04=0,16\left(mol\right)\)

Theo pt (1): \(n_{HCl\left(p.ứ\right)}=0,04y=0,16\Rightarrow y=4\)

Chỉ có Fe₃O₄ là thỏa mãn

`=>` CTHH: \(Fe_3O_4\)

`1.` Ta có: \(\left\{{}\begin{matrix}AB\perp AC\\AB\perp HE\end{matrix}\right.\) `=>AC////HE`

`=>` Tứ giác AEHC là hình thang vuông 

`2.` Xét tam giác vuông HEB, có:  \(\widehat{EBC}=40^o\) \(\Rightarrow\widehat{EHB}=90^o-40^o=50^o\)

Ta có: \(\widehat{EHC}+\widehat{EHB}=180^o\) ( kề bù )

 \(\Rightarrow\widehat{EHC}=180^o-50^o=130^o\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\) ; \(\left(x\ge0\right)\)

\(\Leftrightarrow2x-4\sqrt{x}+\sqrt{x}-2=7\)

\(\Leftrightarrow2x-3\sqrt{x}-9=0\)

\(\Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\)

\(\Leftrightarrow\sqrt{x}-3=0\) ( vì \(2\sqrt{x}+3>0\) )

\(\Leftrightarrow\sqrt{x}=3\)

`<=>x=9`

Vậy \(S=\left\{9\right\}\)

\(\left\{{}\begin{matrix}\sqrt{x+3}+2\sqrt{y+1}=3\left(1\right)\\2\sqrt{x+3}+\sqrt{y+1}=5\left(2\right)\end{matrix}\right.\) ; \(\left(x\ge-3;y\ge-1\right)\)

`<=>`\(\left\{{}\begin{matrix}2\sqrt{x+3}+4\sqrt{y+1}=6\left(1\right)\\2\sqrt{x+3}+\sqrt{y+1}=5\left(2\right)\end{matrix}\right.\)

Lấy \(\left(2\right)-\left(1\right)\), ta được:

\(\left\{{}\begin{matrix}-3\sqrt{y+1}=-1\\\sqrt{x+3}+2\sqrt{y+1}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=\dfrac{1}{3}\\\sqrt{x+3}+2.\dfrac{1}{3}=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+1=\dfrac{1}{9}\\x+3=\dfrac{49}{9}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{8}{9}\\x=\dfrac{22}{9}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(\dfrac{22}{9};-\dfrac{8}{9}\right)\)

áa sai giống tui lúc đầu học:"))

`a.`\(2+\sqrt{2x-1}=x\); \(\left(x\ge2\right)\)

\(\Leftrightarrow\sqrt{2x-1}=x-2\)

\(\Leftrightarrow\left(\sqrt{2x-1}\right)^2=\left(x-2\right)^2\)

\(\Leftrightarrow2x-1=x^2-4x+4\)

\(\Leftrightarrow x^2-6x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\) 

Vậy \(S=\left\{5\right\}\)

`b.`\(\sqrt{x+1}-\sqrt{x-2}=1\); \(\left(x\ge2\right)\)

\(\Leftrightarrow\left(\sqrt{x+1}-\sqrt{x-2}\right)^2=1^2\)

\(\Leftrightarrow x+1+x-2-2\sqrt{\left(x+1\right)\left(x-2\right)}=1\)

\(\Leftrightarrow2x-2=2\sqrt{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow x-1=\sqrt{\left(x+1\right)\left(x-2\right)}\)

\(\Leftrightarrow x^2-2x+1=\left(x+1\right)\left(x-2\right)\)

\(\Leftrightarrow x^2-2x+1=x^2-x-2\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

`c.`\(x^2+2x+\sqrt{2x^2+4x+1}=1\)

\(ĐK:2x^2+4x+1\ge0\)

\(\Leftrightarrow x^2+2x-1+\sqrt{2x^2+4x+1}=0\)

Đặt \(x^2+2x=t\) \(\Rightarrow2x^2+4x=2t\)

Ptr trở thành:

\(t-1+\sqrt{2t+1}=0\)

\(\Leftrightarrow\sqrt{2t+1}=1-t\) \(\left(t\le1\right)\)

\(\Leftrightarrow2t+1=1-2t+t^2\)

\(\Leftrightarrow t^2-4t=0\)

\(\Leftrightarrow t\left(t-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=0\left(tm\right)\\t=4\left(ktm\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\) (tm)

Vậy \(S=\left\{0;-2\right\}\)

\(J=x-6\sqrt{x-1}+2\)

\(J=\left(x+1\right)-2.3\sqrt{x+1}+9-8\)

\(J=\left(\sqrt{x+1}-3\right)^2-8\ge-8\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x+1}-3=0\)

                       \(\Leftrightarrow x=8\)

Vậy \(Min_J=-8\) khi `x=8`