`a.`\(Na_2O+H_2O\rightarrow2NaOH\)

`b.`\(Na_2O+H_2O\rightarrow2NaOH\)

\(Fe_2\left(SO_4\right)_3+6NaOH\rightarrow2Fe\left(OH\right)_3\downarrow+3Na_2SO_4\)

`c.`\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+Na_2SO_4\)

`a.`O: \(K_2O;BaO;MgO;Al_2O_3;\left(FeO;Fe_2O_3\right);H_2O;Ag_2O\)

      Cl: \(KCl;BaCl_2;MgCl_2;AlCl_3;\left(FeCl_2;FeCl_3\right);HCl;AgCl\)

`b.` \(\left(-NO_3\right):NaNO_3;Ca\left(NO_3\right)_2;Zn\left(NO_3\right)_2;Al\left(NO_3\right)_3;Cu\left(NO_3\right)_2;HNO_3\)

\(\left(\equiv PO_4\right)Na_3PO_4;Ca_3\left(PO_4\right)_2;Zn_3\left(PO_4\right)_2;AlPO_4;Cu_3\left(PO_4\right)_2;H_3PO_4\)

\(\left(=CO_3\right)Na_2CO_3;CaCO_3;ZnCO_3;Al_2\left(CO_3\right)_3;CuCO_3;H_2CO_3\)

\(\left(=SO_4\right)Na_2SO_4;CaSO_4;ZnSO_4;Al_2\left(SO_4\right)_3;CuSO_4;H_2SO_4\)

\(\left(-OH\right)NaOH;Ca\left(OH\right)_2;Zn\left(OH\right)_2;Al\left(OH\right)_3;Cu\left(OH\right)_2;H_2O\)

\(y^3=x^3+x^2+x+1\) (1)

Ta có: \(x^2+x+1>0\); \(5x^2+11x+7>0\)

\(\Rightarrow\left(x^3+x^2+x+1\right)-\left(x^2+x+1\right)< x^3+x^2+x+1< \left(x^3+x^2+x+1\right)+\left(5x^2+11x+7\right)=\left(x+2\right)^3\)

\(\Rightarrow x^3< y^3< \left(x+2\right)^3\Rightarrow y^3=\left(x+1\right)^3\) thế vào (1)

\(\Rightarrow\left(x+1\right)^3=x^3+x^2+x+1\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;1\right)\left(-1;0\right)\)

\(2x^6+y^2-2x^3y=320\)

\(\Leftrightarrow x^6+\left(x^6+y^2-2x^3y\right)=320\)

\(\Leftrightarrow\left(x^3\right)^2+\left(x^3-y\right)^2=320\)

Đến đây bạn kẻ bảng là được 

\(\left(320=8^2+16^2;\left(-8\right)^2+\left(-16\right)^2\right)\)

\(\left(2\right)FeCl_3+3NaOH\rightarrow Fe\left(OH\right)_3+3NaCl\)

`(1)Fe_2(SO_4)_3+3BaCl_2->2FeCl_3+3BaSO_4`

`(2)FeCl_3+NaOH->Fe(OH)_3+3NaCl`

`(3)Fe_2(SO_4)_3+6NaOH->2Fe(OH)_3+3Na_2SO_4`

`(4)2Fe(OH)_3+3H_2SO_4->Fe_2(SO_4)_3+6H_2O`

`(5)2Fe(OH)_3->(t^o)Fe_2O_3+3H_2O`

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

\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\dfrac{1}{6}-\dfrac{1}{8}=\left(\dfrac{2}{3}-\dfrac{1}{6}\right)-\dfrac{1}{8}=\dfrac{3}{6}-\dfrac{1}{8}=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)

`a.`\(\widehat{yOz}=180^o-30^o=150^o\)

`b.`\(\widehat{yOz}=180^o-120^o=60^o\)

`c.`\(\widehat{yOz}=180^o-45^o=135^o\)

`d.`\(\widehat{yOz}=180^o-135^o=45^o\)

\(20^{3x}=20^{13}:\left(4.5\right)^4\)

\(\Rightarrow20^{3x}=20^{13}:20^4\)

\(\Rightarrow20^{3x}=20^9\)

`=>3x=9`

`=>x=3`