Question

cho P(x) + (\(3x^2\) - 2) = \(x^3\) + \(3x^2\) - 2x + 2022

tìm P(x)

Answer 1

Ta có : `P(x)+(3x^2-2)=x^3+3x^2-2x+2022`

`->P(x)= x^3+3x^2-2x+2022 -(3x^2-2)`

`= x^3+3x^2-2x+2022 - 3x^2 +2`

`=x^3 +(3x^2-3x^2) -2x+(2022+2)`

`=x^3 -2x+2024`

Answer 2

\(P\left(x\right)=x^3+3x^2-2x+2022-\left(3x^2-2\right)\\ =x^3+3x^2-2x+2022-3x^2+2\\ =x^3+\left(3x^2-3x^2\right)-2x+\left(2022+2\right)\\ =x^3-2x+2024\)

Answer 3

`P(x)+(3x^2-2)=x^3+3x^2-2x+2022`

`-> P(x)= (x^3+3x^2-2x+2022)-(3x^2-2)`

`= x^3+3x^2-2x+2022-3x^2+2`

`= x^3+(3x^2-3x^2)-2x+(2022+2)`

`= x^3-2x+2024`

Vậy, `P(x)=x^3-2x+2024`

Answer 4

\(P\left(x\right)+\left(3x^2-2\right)=x^3+3x^2-2x+2022\)

\(\Rightarrow P\left(x\right)=x^3+3x^2-2x+2022-\left(3x^2-2\right)\)

\(\Rightarrow P\left(x\right)=x^3+3x^2-2x+2022-3x^2+2\)

\(\Rightarrow P\left(x\right)=x^3+\left(3x^2-3x^2\right)-2x+\left(2022+2\right)\)

\(\Rightarrow P\left(x\right)=x^3-2x+2024\)

Vậy \(P\left(x\right)=x^3-2x+2024\)