`a.` \(2K+2HCl\rightarrow2KCl+H_2\)

\(2K+H_2SO_4\rightarrow K_2SO_4+H_2\)

`6K+2H_3PO_4->2K_3PO_4+3H_2`

`Fe+2HCl->FeCl_2+H_2`

`Fe+H_2SO_4->FeSO_4+H_2`

`3Fe+2H_3PO_4->Fe_3(PO_4)_2+3H_2`

`Zn+2HCl->ZnCl_2+H_2`

`Zn+H_2SO_4->ZnSO_4+H_2`

`3Zn+2H_3PO_4->Zn_3(PO_4)_2+3H_2`

`Ag+HCl` ( ko p.ứ )

`Ag+H_2SO_4` ( ko p.ứ )

`Ag+H_3PO_4` ( ko p.ứ )

`Ba+2HCl->BaCl_2+H_2`

`Ba+H_2SO_4->BaSO_4+H_2`

`3Ba+2H_3PO_4->Ba_3(PO_4)_2+3H_2`

`b.` HNO₃:

`K_2O+2HNO_3->2KNO_3+H_2O`

`KOH+HNO_3->KNO_3+H_2O`

`Al(OH)_3+3HNO_3->Al(NO_3)_3+3H_2O`

`CaO+HNO_3->Ca(NO_3)_2+H_2O`

`Fe_2O_3+6HNO_3->2Fe(NO_3)_3+3H_2O`

`CuO+2HNO_3->Cu(NO_3)_2+H_2O`

`Ca(OH)_2+HNO_3->Ca(NO_3)_2+H_2O`

`Al_2O_3+6HNO_3->2Al(NO_3)_3+3H_2O`

H₂SO₄:

`K_2O+H_2SO_4->K_2SO_4+H_2O`

`2KOH+H_2SO_4->K_2SO_4+2H_2O`

`2Al(OH)_3+3H_2SO_4->Al_2(SO_4)_3+6H_2O`

`CaO+H_2SO_4->CaSO_4+H_2O`

`Fe_2O_3+3H_2SO_4->Fe_2(SO_4)_3+3H_2O`

`CuO+H_2SO_4->CuSO_4+H_2O`

`Ca(OH)_2+H_2SO_4->CaSO_4+2H_2O`

`Al_2O_3+3H_2SO_4->Al_2(SO_4)_3+3H_2O`

`c.` 

\(CO_2+H_2O\rightarrow H_2CO_3\)

`P_2O_5+3H_2O->2H_3PO_4`

`N_2O_5+H_2O->2HNO_3`

`SiO_2+H_2O` ( ko p.ứ )

`CaO+H_2O->Ca(OH)_2`

`CuO+H_2O` ( ko p.ứ )

Bạn sửa đề à??

\(\sqrt[3]{9-x}+\sqrt{x+3}=4\)

\(ĐK:x\ge-3\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{9-x}=t\\\sqrt{x+3}=p\end{matrix}\right.\); \(p\ge0\) \(\Rightarrow\left\{{}\begin{matrix}t^3=9-x\\p^2=x+3\end{matrix}\right.\) \(\Leftrightarrow t^3+p^2=12\)

Ta có hpt: \(\left\{{}\begin{matrix}t+p=4\\t^3+p^2=12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}p=4-t\\t^3+p^2=12\left(2\right)\end{matrix}\right.\)

\(\left(2\right)\Leftrightarrow t^3+\left(4-t\right)^2=12\)

\(\Leftrightarrow t^3+t^2-8t+16=12\)

\(\Leftrightarrow t^3+t^2-8t+4=0\)

\(\Leftrightarrow t^3-2t^2+3t^2-8t+4=0\)

\(\Leftrightarrow t^2\left(t-2\right)+\left(3t^2-6t-2t+4\right)=0\)

\(\Leftrightarrow t^2\left(t-2\right)+3t\left(t-2\right)-2\left(t-2\right)=0\)

\(\Leftrightarrow\left(t-2\right)\left(t^2+3t-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t^2+3t-2=0\left(3\right)\end{matrix}\right.\)

\(\left(3\right)t^2+3t-2=0\)

\(\Delta=3^2-4.\left(-2\right)=17>0\)

\(\rightarrow\left[{}\begin{matrix}t=\dfrac{-3+\sqrt{17}}{2}\\t=\dfrac{-3-\sqrt{17}}{2}\end{matrix}\right.\)

`@` TH1: \(t=2\Rightarrow p=4-2=2\left(tm\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{9-x}=2\\\sqrt{x+3}=2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}9-x=8\\x+3=4\end{matrix}\right.\) \(\Leftrightarrow x=1\left(tm\right)\)

`@`TH2:  \(t=\dfrac{-3\pm\sqrt{17}}{2}\) \(\Rightarrow p=4-\dfrac{-3\pm\sqrt{17}}{2}=\dfrac{11\pm\sqrt{17}}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt[3]{9-x}=\dfrac{-3\pm\sqrt{17}}{2}\\\sqrt{x+3}=\dfrac{11\pm\sqrt{17}}{2}\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{63\pm11\sqrt{17}}{2}\left(tm\right)\)

Vậy \(S=\left\{1;\dfrac{63\pm11\sqrt{17}}{2}\right\}\)

\(\dfrac{x}{x+1}-2\sqrt{\dfrac{x+1}{x}}=3\)

\(ĐK:\sqrt{\dfrac{x+1}{x}}\ge0\)

Đặt \(\sqrt{\dfrac{x+1}{x}}=t;t\ge0\)

\(\Leftrightarrow\dfrac{x+1}{x}=t^2\) \(\Leftrightarrow\dfrac{x}{x+1}=\dfrac{1}{t^2}\)

Ptr trở thành:

\(\Leftrightarrow\dfrac{1}{t^2}-2t=3\)

\(\Leftrightarrow\dfrac{1-2t^3}{t^2}=3\)

\(\Leftrightarrow1-2t^3=3t^2\)

\(\Leftrightarrow2t^3+3t^2-1=0\)

\(\Leftrightarrow2t^3+2t^2+t^2-1=0\)

\(\Leftrightarrow2t^2\left(t+1\right)+\left(t-1\right)\left(t+1\right)=0\)

\(\Leftrightarrow\left(t+1\right)\left(2t^2+t-1\right)=0\)

\(\Leftrightarrow2t^2+t-1=0\) ( vì \(t+1>0\) )

\(\Leftrightarrow\left[{}\begin{matrix}t=-1\left(ktm\right)\\t=\dfrac{1}{2}\left(tm\right)\end{matrix}\right.\) ( Vi-ét )

`<=>`\(\sqrt{\dfrac{x+1}{x}}=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x}=\dfrac{1}{4}\)

\(\Leftrightarrow4x+4=x\)

\(\Leftrightarrow3x+4=0\)

\(\Leftrightarrow x=-\dfrac{4}{3}\left(tm\right)\)

\(\dfrac{260abc}{626}=abc\)

\(\Rightarrow260abc=626abc\)

\(\Rightarrow626abc-260abc=0\)

\(\Rightarrow366abc=0\)

\(\Rightarrow abc=0\)

`@`\(2023+abc=2023+0=2023\)

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