Question

Cho đa thức P(x)=ax^2+bx+c.

Hãy xác định các số a,b,c biết rằng P(0)=1,P(1)=-1,P(2)=2

Answer 1

`P(0)=1`

`<=>a.0+b.0+c=1`

`<=>c=1`

`P(1)=-1`

`<=>a.1+b.1+c=-1`

`<=>a+b+1=-1`

`<=>a+b=-2<=>a=-2-b(1)`

`P(2)=2`

`<=>a.4+b.2+c=2`

`<=>4a+2b=1(2)`

Thay `a=-2-b` từ (1) vào (2) ta có:

`=-8-4b+2b=1`

`<=>-2b=7`

`<=>b=-7/2`

`<=>a=-1/2`

Vậy `a=-1/2,b=-7/2,c=1`

Answer 2

\(P\left(0\right)=c=1\\ P\left(1\right)=a+b+c=-1\\ \Rightarrow a+b=-2\\ \Rightarrow a=-2-b\\ P\left(2\right)=4a+2b+c=2\\ \Rightarrow4\left(-2-b\right)+2b=1\\ \Rightarrow-8-4b+2b=1\\ \Rightarrow-2b=9\\ \Rightarrow b=\dfrac{-9}{2}\\ \Rightarrow a=-2+\dfrac{9}{2}=\dfrac{5}{2}\)

Vậy \(\left(a;b;c\right)=\left(\dfrac{5}{2};\dfrac{-9}{2};1\right)\)