b1 :tính: (3x mũ 4 – 8x mũ 3- 10x mũ 2+8x-5):(3x mũ 2 – 2x + 1) 11/09/2021 Bởi Natalia b1 :tính: (3x mũ 4 – 8x mũ 3- 10x mũ 2+8x-5):(3x mũ 2 – 2x + 1)
`3x^4-8x^3-10x^2+8x-5` `=3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5` `=(3x^4-2x^3+x^2)-(6x^3-4x^2+2x)-(15x^2-10x+5)` `=x^2(3x^2-2x+1)-2x(3x^2-2x+1)-5(3x^2-2x+1)` `=(3x^2-2x+1)(x^2-2x-5).` Có: `(3x^4-8x^3-10x^2+8x-5):(3x^2-2x+1)` `⇔(3x^2-2x+1)(x^2-2x-5):(3x^2-2x+1)=x^2-2x-5.` Vậy `(3x^4-8x^3-10x^2+8x-5):(3x^2-2x+1)=x^2-2x-5.` Bình luận
Đặt A : B = ($3x^{4}$ – $8x^{3}$ – $10x^{2}$ + 8x -5) : ( $3x^{2}$ – 2x + 1 ) ( 1 ) ⇒ A = $3x^{4}$ – $2x^{3}$ + $x^{2}$ – $6x^{3}$ + $4x^{2}$ – 2x – $15^{2}$ + 10x – 5 ⇒ A = ( $3x^{4}$ – $2x^{3}$ + $x^{2}$ ) – ( $6x^{3}$ – $4x^{2}$ + 2x) – ( $15x^{2}$ – 10x + 5) ⇒ A = $x^{2}$ . ( $3x^{2}$ – 2x + 1 ) – 2x. ( $3x^{2}$ – 2x + 1 ) – 5 . ( $3x^{2}$ – 2x + 1 ) ⇒ A = ( $3x^{2}$ – 2x + 1 ) . ( $x^{2}$ – 2x – 5 ) ( 2 ) Từ (1) và (2) ra ta có: A : B = ( $3x^{2}$ – 2x + 1 ) . ( $x^{2}$ – 2x – 5 ) : ( $3x^{2}$ – 2x + 1 ) = $x^{2}$ – 2x – 5 Bình luận
`3x^4-8x^3-10x^2+8x-5`
`=3x^4-2x^3+x^2-6x^3+4x^2-2x-15x^2+10x-5`
`=(3x^4-2x^3+x^2)-(6x^3-4x^2+2x)-(15x^2-10x+5)`
`=x^2(3x^2-2x+1)-2x(3x^2-2x+1)-5(3x^2-2x+1)`
`=(3x^2-2x+1)(x^2-2x-5).`
Có: `(3x^4-8x^3-10x^2+8x-5):(3x^2-2x+1)`
`⇔(3x^2-2x+1)(x^2-2x-5):(3x^2-2x+1)=x^2-2x-5.`
Vậy `(3x^4-8x^3-10x^2+8x-5):(3x^2-2x+1)=x^2-2x-5.`
Đặt A : B = ($3x^{4}$ – $8x^{3}$ – $10x^{2}$ + 8x -5) : ( $3x^{2}$ – 2x + 1 ) ( 1 )
⇒ A = $3x^{4}$ – $2x^{3}$ + $x^{2}$ – $6x^{3}$ + $4x^{2}$ – 2x – $15^{2}$ + 10x – 5
⇒ A = ( $3x^{4}$ – $2x^{3}$ + $x^{2}$ ) – ( $6x^{3}$ – $4x^{2}$ + 2x) – ( $15x^{2}$ – 10x + 5)
⇒ A = $x^{2}$ . ( $3x^{2}$ – 2x + 1 ) – 2x. ( $3x^{2}$ – 2x + 1 ) – 5 . ( $3x^{2}$ – 2x + 1 )
⇒ A = ( $3x^{2}$ – 2x + 1 ) . ( $x^{2}$ – 2x – 5 ) ( 2 )
Từ (1) và (2) ra ta có:
A : B = ( $3x^{2}$ – 2x + 1 ) . ( $x^{2}$ – 2x – 5 ) : ( $3x^{2}$ – 2x + 1 )
= $x^{2}$ – 2x – 5