Đáp án+Giải thích các bước giải:
Bài 3:
`a) (a+b)^3+(a-b)^3-6ab^2`
`=a^3+3a^2b+3ab^2+b^3+a^3-3a^2b+3ab^2-b^3-6ab^2`
`=(a^3+a^3)+(3a^2b-3a^2b)+(3ab^2+3ab^2-6ab^2)+(b^3-b^3)`
`=2a^3`
`b) (a+b)^3-(a-b)^3-6a^2b`
`=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-6a^2b`
`=(a^3-a^3)+(3ab^2-3ab^2)+(3a^2b+3a^2b-6a^2b)+(b^3+b^3)`
`=2b^3`
`c) (x+2)^3+(x-2)^3-2x(x^2+12)`
`=x^3+6x^2+12x+8+x^3-6x^2+12x-8-2x^3-24x`
`=(x^3+x^3-2x^3)+(6x^2-6x^2)+(12x+12x-24x)+(8-8)`
`=0`
`d) (x-1)^3-(x+1)^3+6(x+1)(x-1)`
`=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6(x^2-1)`
`=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x^2-6`
`=(x^3-x^3)-(3x^2+3x^2-6x^2)+(3x-3x)-(1+1+6)`
`=-8`
Bài 4:
`a) A=x^2+5x+7`
`=x^2+2.x. 5/2 + (25)/4 +3/4`
`=(x+5/2)^2+3/4`
Vì `(x+5/2)^2 ≥ 0 ∀ x ∈RR`
nên `(x+5/2)^2+3/4≥ 3/4 ∀ x ∈RR`
hay `A ≥ 3/4 ∀ x ∈RR`
`A_(min) <=> x+5/2=0`
`<=> x= -5/2`
Vậy `A_(min)=3/4` khi `x=-5/2`
`b) B=6x^2-x^2-5`
`= -(x^2-6x+5)`
`=-(x^2-2.x.3+9-4)`
`=-(x-3)^2+4`
Vì `-(x-3)^2≤0 ∀ x ∈RR`
nên `-(x-3)^2+4 ≤ 4∀x∈RR`
hay `B≤ 4∀x∈RR`
`B_(max) <=> x-3=0`
`<=> x = 3`
Vậy `B_(max)=4` khi `x=3`
