Xet tam giac BDC va tam giac CEB ta co 

^BDC = ^CEB = 90⁰

BC _ chung 

^BCD = ^CBE ( gt ) 

=> tam giac BDC = tam giac CEB ( ch - gn ) 

=> ^DBC = ^ECB ( 2 goc tuong ung ) 

Ta co ^B - ^DBC = ^ABD 

^C - ^ECB = ^ACE 

=> ^ABD = ^ACE 

Xet tam giac IBE va tam giac ICD 

^ABD = ^ACE ( cmt )

^BIE = ^CID ( doi dinh ) 

^BEI = ^IDC = 90⁰

Vay tam giac IBE = tam giac ICD (g.g.g) 

c, Do BD vuong AC => BD la duong cao 

CE vuong BA => CE la duong cao 

ma BD giao CE = I => I la truc tam 

=> AI la duong cao thu 3 

=> AI vuong BC 

`a,A(x) =2x^3 -x^4 +2x-4+3x^2 -2x^3+x^4`

`= ( 2x^3-2x^3) +(-x^4+x^4) + 2x -4+3x^2`

`= 0+0+ 2x -4+3x^2`

`= 3x^2 +2x-4`

`b, M(x)=A(x)+B(x)`

`M(x)= 3x^2 +2x-4 + x-2`

`= 3x^2 + 3x-6`

`b,  N(x) = A(x) - B(x)`

`N(x)=  3x^2 +2x-4 -(x-2)`

`= 3x^2 +2x-4 -x+2`

`= 3x^2 + x -2`

`c,` Ta có :

`x-2=0`

`=> x=0+2`

`=>x=2`