Đáp án:
Giải thích các bước giải:
a) `3x(x-4)-x(5+3x)=-34`
`⇔ 3x^2-12x-5x-3x^2=-34`
`⇔ -17x=-34`
`⇔ x=(-34):(-17)`
`⇔ x=2`
Vậy `S={2}`
b) `4x^2-12x+9=0`
`⇔ (2x)^2-2.2.3x+(3)^2=0`
`⇔ (2x-3)^2=0`
`⇔ 2x-3=0`
`⇔ x=3/2`
Vậy `S={3/2}`
Bài 7:
a) `A=4x^2-12x+2015`
`A=4x^2-12x+9+2006`
`A=(2x)^2-2.2.3x+(3)^2+2006`
`A=(2x-3)^2+2006`
Ta có: `(2x-3)^2 \ge 0 ∀x`
`⇒ (2x-3)^2+2006 \ge 2006 \forall x`
Vậy `A_{min}=2006` khi `2x-3=0⇔x=3/2`
b) `B=5x-x^2`
`B=-(x^2-2 . 5/2 x+25/4-25/4)`
`B=-[(x-5/2)^2+25/4]`
`B=-(x-5/2)^2-25/4`
Ta có: `-(x-5/2)^2 \le 0 \forall x`
`⇒ -(x-5/2)^2-25/4 \le -25/4 \forall x`
Vậy `B_{max}=-25/4` khi `x-5/2=0⇔ x=5/2`
