Bài 4. Cho P(x)+Q(x)=5x^2-4x+1 và P(x)-Q(x)=x^2+2x-5 Tìm p(x) và Q(x) 12/07/2021 Bởi Kennedy Bài 4. Cho P(x)+Q(x)=5x^2-4x+1 và P(x)-Q(x)=x^2+2x-5 Tìm p(x) và Q(x)
Ta có:`P(x)+Q(x)=5x²-4x+1` `⇒P(x)=5x²-4x+1-Q(x)` Ta có:`P(x)-Q(x)=x²+2x-5` `⇒5x²-4x+1-Q(x)-Q(x)=x²+2x-5` `⇒5x²-4x+1-2Q(x)=x²+2x-5` `⇒-2Q(x)=x²+2x-5-(5x²-4x+1)` `⇒-2Q(x)=x²+2x-5-5x²+4x-1` `⇒-2Q(x)=-4x²+6x-6` `⇒Q(x)=(-4x²+6x-6):(-2)` `⇒Q(x)=2x²-3x+3` Ta có:`P(x)=5x²-4x+1-Q(x)` `⇒P(x)=5x²-4x+1-(2x²-3x+3)` `⇒P(x)=5x²-4x+1-2x²+3x-3` `⇒P(x)=3x²-x-2` Vậy `P(x)=3x²-x-2` và `Q(x)=2x²-3x+3` Bình luận
$\text{Ta có:}$ $\text{P(x)+Q(x)=5x² -4x+1}$ $\text{P(x)-Q(x)=x²+2x-5}$ $\text{⇒ P(x)+Q(x)-[P(x)-Q(x)]=(5x² -4x+1)-(x²+2x-5)}$ $\text{⇔ P(x)+Q(x)-P(x)+Q(x)=5x² -4x+1-x²-2x+5}$ $\text{⇔ 2.Q(x)=4x²-6x+6}$ $\text{⇔ Q(x)=(4x²-6x+6):2}$ $\text{⇔ Q(x)=2x²-3x+3}$ $\text{P(x)-Q(x)=x²+2x-5}$ $\text{⇒ P(x)=x²+2x-5+Q(x)}$ $\text{hay P(x)=x²+2x-5+2x²-3x+3}$ $\text{⇔ P(x)=3x²-x-2}$ Bình luận
Ta có:`P(x)+Q(x)=5x²-4x+1`
`⇒P(x)=5x²-4x+1-Q(x)`
Ta có:`P(x)-Q(x)=x²+2x-5`
`⇒5x²-4x+1-Q(x)-Q(x)=x²+2x-5`
`⇒5x²-4x+1-2Q(x)=x²+2x-5`
`⇒-2Q(x)=x²+2x-5-(5x²-4x+1)`
`⇒-2Q(x)=x²+2x-5-5x²+4x-1`
`⇒-2Q(x)=-4x²+6x-6`
`⇒Q(x)=(-4x²+6x-6):(-2)`
`⇒Q(x)=2x²-3x+3`
Ta có:`P(x)=5x²-4x+1-Q(x)`
`⇒P(x)=5x²-4x+1-(2x²-3x+3)`
`⇒P(x)=5x²-4x+1-2x²+3x-3`
`⇒P(x)=3x²-x-2`
Vậy `P(x)=3x²-x-2` và `Q(x)=2x²-3x+3`
$\text{Ta có:}$
$\text{P(x)+Q(x)=5x² -4x+1}$
$\text{P(x)-Q(x)=x²+2x-5}$
$\text{⇒ P(x)+Q(x)-[P(x)-Q(x)]=(5x² -4x+1)-(x²+2x-5)}$
$\text{⇔ P(x)+Q(x)-P(x)+Q(x)=5x² -4x+1-x²-2x+5}$
$\text{⇔ 2.Q(x)=4x²-6x+6}$
$\text{⇔ Q(x)=(4x²-6x+6):2}$
$\text{⇔ Q(x)=2x²-3x+3}$
$\text{P(x)-Q(x)=x²+2x-5}$
$\text{⇒ P(x)=x²+2x-5+Q(x)}$
$\text{hay P(x)=x²+2x-5+2x²-3x+3}$
$\text{⇔ P(x)=3x²-x-2}$