\(\dfrac{-x^{2n}y^{2n-1}}{\dfrac{1}{5}x^{2n-1}y^{2n-4}}=-5x^{-1}y^3=-\dfrac{1}{5}xy^3\)

Bài 1 : 

\(1) Fe_2O_3 + 6HCl \to 2FeCl_3 + 3H_2O\\ (2) FeCl_3 + 3AgNO_3 \to 3AgCl + Fe(NO_3)_3\\ (3)Fe(NO_3)_3 + 3KOH \to Fe(OH)_3 + 3KNO_3\\ (4)2Fe(OH)_3 +3 H_2SO_4 \to Fe_2(SO_4)_3 + 6H_2O\\ (5)Fe_2(SO_4)_3 + 6NaOH \to 2Fe(OH)_3 + 3Na_2SO_4\\ (6)2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ b) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ Al_2O_3 + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2O\\ Al_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2AlCl_3\\ AlCl_3 + 3AgNO_3 \to 3AgCl + Al(NO_3)_3\\ Al(NO_3)_3 + 3KOH \to Al(OH)_3 + 3KNO_3\\ \)

\(2Al(OH)_3 \xrightarrow{t^o} Al_2O_3 + 3H_2O\\ c) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ Fe_2(SO_4)_3 + 3BaCl_2 \to 3BaSO_4 + 2FeCl_3\\ FeCl_3 + 3KOH \to Fe(OH)_3 + 3KCl\\ 2Fe(OH)_3 \xrightarrow{t^o} Fe_2O_3 + 3H_2O\\ Fe_2O_3 + 3H_2 \xrightarrow{t^o} 2Fe + 3H_2O\\ Fe + 2HCl \to FeCl_2 + H_2\\ 2FeCl_2 + Cl_2 \to 2FeCl_3\\ d) CuCl_2 + 2AgNO_3 \to 2AgCl + Cu(NO_3)_2\\ Cu(NO_3)_2 + 2KOH \to Cu(OH)_2 +2KNO_3\\ Cu(OH)_2 + H_2SO_4 \to CuSO_4 + 2H_2O\\ CuSO_4 + Fe \to FeSO_4 + Cu\\ \)

\(2Cu + O_2 \to 2CuO\\ CuO + H_2 \xrightarrow{t^o} Cu + H_2O\)

Bài 2 : 

$(1) 2SO_2 + O_2 \xrightarrow{t^o,xt} 2SO_3$
$(2) SO_3 + H_2O \to H_2SO_4$
$(3) BaCl_2 + H_2SO_4 \to BaSO_4 + 2HCl$
$(4) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O$
$(5) 2NaOH + SO_2 \to Na_2SO_3 + H_2O$
$(6) Na_2SO_3 + 2HCl \to 2NaCl + SO_2 + H_2O$

A B C E H I

a/ Xét tg vuông ABH có

\(\widehat{B}=90^o-\widehat{BAH}=90^o-2\widehat{C}\)

\(\widehat{A}=180^0-\widehat{B}-\widehat{C}=180^o-90^o+2\widehat{C}-\widehat{C}=90^o+\widehat{C}\)

Ta có

\(\widehat{BAI}=\widehat{HAI}=\dfrac{\widehat{BAH}}{2}=\dfrac{2\widehat{C}}{2}=\widehat{C}\)

Ta có

\(\widehat{A}=\widehat{BAI}+\widehat{EAI}=\widehat{C}+\widehat{EAI}=90^o+\widehat{C}\Rightarrow\widehat{EAI}=90^o\) (1)

Xét tg ABI có

\(\widehat{AIE}=\widehat{ABI}+\widehat{BAI}=\dfrac{\widehat{B}}{2}+\widehat{C}\) (trong tg góc ngoài bằng tổng 2 góc trong không kề với nó)

Xét tg BCE có

\(\widehat{AEI}=\widehat{EBC}+\widehat{C}=\dfrac{\widehat{B}}{2}+\widehat{C}\)

\(\Rightarrow\widehat{AIE}=\widehat{AEI}\) (2)

Từ (1) và (2) => tg AIE vuông cân tại A \(\Rightarrow\widehat{AIE}=\widehat{AEI}=45^o\)

b/

Xét tg ABH

Nối HI => HI là phân giác của \(\widehat{AHB}\) (trong tg 3 đường phân giác đồng quy) \(\Rightarrow\widehat{AHI}=\widehat{BHI}=45^o\)

Ta có \(\widehat{AIE}=\widehat{AEI}=45^o\) (cmt)

\(\Rightarrow\widehat{AEI}=\widehat{AHI}=45^o\)

=> tứ giác AEHI là tứ giác nội tiếp

\(\Rightarrow\widehat{AHE}=\widehat{AIE}=45^o\) (góc nội tiếp cùng chắn cung AE)

\(\Rightarrow\widehat{CHE}=\widehat{AHC}-\widehat{AHE}=90^o-45^o=45^o\)

\(\Rightarrow\widehat{AHE}=\widehat{CHE}=45^o\) => HE là phân giác của \(\widehat{AHC}\)

đk x khác -2 ; 2 

\(A=\dfrac{2x+4+4x-8-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

`(4x^2-2x+1).(2x+1)+(2x-3)^3`

`=8x^3+1+8x^3 -36x^2 +54x-27`

`=(8x^3+8x^3)+(1-27)-36x^2 +54x`

`=16x^3 - 26 -36x^2+54x`

\(=8x^3+1+8x^3-3.8x^3.3+3.2x.27-27\)

\(=16x^3-26-72x^3+162x\)

1, \(=4x^2-12x+9\)

2, \(=16y^4+8y^2+1\)

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

4, \(=8x^6-3.4x^4.3y+3.2x^2.9y^2+27y^3=8x^6-36x^4y+54x^2y^2+27y^3\)

6, \(=\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)