cho đa thức A(x)= ax ² +bx+c biết b=5a+c CMR:A(+1)×A(-3) ≤ 0 19/10/2021 Bởi Arianna cho đa thức A(x)= ax ² +bx+c biết b=5a+c CMR:A(+1)×A(-3) ≤ 0
`A(x)=ax^2+bx+c` `=ax^2+(5a+c)x+c` `=ax^2+5ax+cx+x` `=ax(x+5)+x(c+1)` `A(1).(A(-3)=[a.1(1+5)+1.(c+1)][a(-3)(-3+5)+(-3)(c+1)]` `=(6a+c+1)(-6a+-3c-3)` `=-(6a+c+1)^2` Ta có :`(6a+c+1)^2 ≥ 0 ∀ x` `⇒-(6a+c+1)^2 ≤ 0 ∀ x` `⇔A(1).A(-3)≤0` Bình luận
Giải thích các bước giải: `A(x)= ax^2 +bx+c=ax^2+(5a+c).x+c` `=ax^2+5ax+cx+c=ax(x+5)+c(x+1)` Ta có: `A(1)=a(1+5)+c(1+1)=6a+2c` `A(-3)=-3a(-3+5)+c(-3+1)=-3a.2+c.(-2)=-6a-2c=-(6a+2c)` `=>A(1).A(3)=-(6a+2c).(6a+2c)=-(6a+2c)^2` Mà `(6a+2c)^2>=0` `=>-(6a+2c)^2<=0` `=>A(1).A(-3)<=0` Vậy `A(1).A(-3)<=0`. Bình luận
`A(x)=ax^2+bx+c`
`=ax^2+(5a+c)x+c`
`=ax^2+5ax+cx+x`
`=ax(x+5)+x(c+1)`
`A(1).(A(-3)=[a.1(1+5)+1.(c+1)][a(-3)(-3+5)+(-3)(c+1)]`
`=(6a+c+1)(-6a+-3c-3)`
`=-(6a+c+1)^2`
Ta có :`(6a+c+1)^2 ≥ 0 ∀ x`
`⇒-(6a+c+1)^2 ≤ 0 ∀ x`
`⇔A(1).A(-3)≤0`
Giải thích các bước giải:
`A(x)= ax^2 +bx+c=ax^2+(5a+c).x+c`
`=ax^2+5ax+cx+c=ax(x+5)+c(x+1)`
Ta có:
`A(1)=a(1+5)+c(1+1)=6a+2c`
`A(-3)=-3a(-3+5)+c(-3+1)=-3a.2+c.(-2)=-6a-2c=-(6a+2c)`
`=>A(1).A(3)=-(6a+2c).(6a+2c)=-(6a+2c)^2`
Mà `(6a+2c)^2>=0`
`=>-(6a+2c)^2<=0`
`=>A(1).A(-3)<=0`
Vậy `A(1).A(-3)<=0`.