Đáp án:
Giải thích các bước giải:
e) `E=(x− 2)(x + 4)`
`E=x^2+4x-2x-8`
`E=x^2+2x-8`
`E=x^2+2x+1-9`
`E=(x+1)^2-9`
Ta có: `(x+1)^2 \ge 0 \forall x`
`⇒ (x+1)^2-9 \ge -9 \forall x`
Vậy `E_{min}=-9` khi `x-1=0⇔x=-1`
f) `F = (2x − 1)(2x + 5)`
`F=4x^2+10x-2x-5`
`F=4x^2+8x-5`
`F=4x^2+8x+4-9`
`F=4(x+1)^2-9`
Ta có: `(x+1)^2 \ge 0 \forall x`
`⇒ 4(x+1)^2-9 \ge -9 \forall x`
Vậy `F_{min}=-9` khi `x+1=0⇔x=-1`
g) `G= (3x − 2)(x + 4)`
`G=3x^2+12x-2x-8`
`G=3x^2+10x-8`
`G=3(x^2+10/3 x-8/3)`
`G=3(x^2+2 . 5/3 x+25/9-49/9)`
`G=3[(x+5/3)^2-49/9]`
`G=3(x+5/3)^2-49/3`
Ta có: `(x+5/3)^2 \ge 0 \forall x`
`⇒ 3(x+5/3)^2 \ge 0 \forall x`
`⇒ 3(x+5/3)^2-49/3 \ge -49/3 \forall x`
Vậy `G_{min}=-49/3` khi `x+5/3=0⇔x=-5/3`
