Câu 2 trang 24 SBT môn Toán 8 tập 1: Dùng định nghĩa hai phân thức bằng nhau, hãy tìm đa thức A trong mỗi...

Dùng định nghĩa hai phân thức bằng nhau, hãy tìm đa thức A trong mỗi đẳng thức sau:

a. ({A over {2x – 1}} = {{6{x^2} + 3x} over {4{x^2} – 1}})

b. ({{4{x^2} – 3x – 7} over A} = {{4x – 7} over {2x + 3}})

c. ({{4{x^2} – 7x + 3} over {{x^2} – 1}} = {A over {{x^2} + 2x + 1}})

d. ({{{x^2} – 2x} over {2{x^2} – 3x – 2}} = {{{x^2} + 2x} over A})

Giải:

a. ({A over {2x – 1}} = {{6{x^2} + 3x} over {4{x^2} – 1}})

( Rightarrow Aleft( {4{x^2} – 1} ight) = left( {2x – 1} ight).left( {6{x^2} + 3x} ight))

( Rightarrow Aleft( {2x – 1} ight)left( {2x + 1} ight) = left( {2x – 1} ight).3xleft( {2x + 1} ight))

( Rightarrow A = 3x)

Ta có: ({{3x} over {2x – 1}} = {{6{x^2} + 3x} over {4{x^2} – 1}})

b. ({{4{x^2} – 3x – 7} over A} = {{4x – 7} over {2x + 3}})

 (eqalign{  &  Rightarrow left( {4{x^2} – 3x – 7} ight)left( {2x + 3} ight) = Aleft( {4x – 7} ight)  cr  &  Rightarrow left( {4{x^2} + 4x – 7x – 7} ight)left( {2x + 3} ight) = Aleft( {4x – 7} ight)  cr  &  Rightarrow left[ {4xleft( {x + 1} ight) – 7left( {x + 1} ight)} ight]left( {2x + 3} ight) = Aleft( {4x – 7} ight)  cr  &  Rightarrow left( {x – 1} ight)left( {4x – 7} ight)left( {2x + 3} ight) = Aleft( {4x – 7} ight)  cr  &  Rightarrow A = left( {x + 1} ight)left( {2x + 3} ight) = 2{x^2} + 3x + 2x + 3 = 2{x^2} + 5x + 3 cr} )

Ta có: ({{4{x^2} – 3x – 7} over {2{x^2} + 5x + 3}} = {{4x – 7} over {2x + 3}})

c. ({{4{x^2} – 7x + 3} over {{x^2} – 1}} = {A over {{x^2} + 2x + 1}})    

(eqalign{  &  Rightarrow left( {4{x^2} – 7x + 3} ight).left( {{x^2} + 2x + 1} ight) = A.left( {{x^2} – 1} ight)left( {{pi  over 2} – heta } ight)  cr  &  Rightarrow left( {4{x^2} – 4x – 3x + 3} ight).{left( {x + 1} ight)^2} = Aleft( {x + 1} ight)left( {x – 1} ight)  cr  &  Rightarrow left[ {4xleft( {x – 1} ight) – 3left( {x – 1} ight)} ight].{left( {x + 1} ight)^2} = Aleft( {x + 1} ight)left( {x – 1} ight)  cr  &  Rightarrow left( {x – 1} ight)left( {4x – 3} ight){left( {x + 1} ight)^2} = Aleft( {x + 1} ight)left( {x – 1} ight)  cr  &  Rightarrow A = left( {4x – 3} ight)left( {x + 1} ight) = 4{x^2} + 4x – 3x – 3 = 4{x^2} + x – 3 cr} )

Ta có:    ({{4{x^2} – 7x + 3} over {{x^2} – 1}} = {{4{x^2} + x – 3} over {{x^2} + 2x + 1}})

d. ({{{x^2} – 2x} over {2{x^2} – 3x – 2}} = {{{x^2} + 2x} over A})    

(eqalign{  &  Rightarrow left( {{x^2} – 2x} ight).A = left( {2{x^2} – 3x – 2} ight)left( {{x^2} + 2x} ight)  cr  &  Rightarrow xleft( {x – 2} ight).A = left( {2{x^2} – 4x + x – 2} ight).xleft( {x + 2} ight)  cr  &  Rightarrow xleft( {x – 2} ight).A = left[ {2xleft( {x – 2} ight) + left( {x – 2} ight)} ight].xleft( {x + 2} ight)  cr  &  Rightarrow xleft( {x – 2} ight).A = left( {2x + 1} ight)left( {x – 2} ight).x.left( {x + 2} ight)  cr &  Rightarrow A = left( {2x + 1} ight)left( {x + 2} ight) = 2{x^2} + 4x + x + 2 = 2{x^2} + 5x + 2 cr} )

Ta có : ({{{x^2} – 2x} over {2{x^2} – 3x – 2}} = {{{x^2} + 2x} over {{x^2} + 2x + 1}})