`#Sad`

`a)`

`(x+1)^3+(x-2)^3-2x^2(x-1,5) = 3`

`⇔ x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2 = 3`

`⇔ (x^3+x^3-2x^3)+(-6x^2+3x^2+3x^2)+(3x+12x)+(1-8) = 3`

`⇔ 15x-7 = 3`

`⇔ 15x = 10`

`⇔ x = 2/3`

`\text{Vậy}` `x=2/3`

`\text{→Áp dụng:}`

`+)` `(A+B)^3 = A^3+3A^2B+3AB^2+B^3`

`+)` `(A-B)^3 = A^3-3A^2B+3AB^2-B^3`

`b)`

`(x+2)(x^2-2x+4)(x-2)(x^2+2x+4) = -65`

`⇔ [(x+2)(x^2-2x+4)][(x-2)(x^2+2x+4)] = -65`

`⇔ (x^3+8)(x^3-8) = -65`

`⇔ x^6-8x^3+8x^3-64 = -65`

`⇔ x^6 = -1`

$\text{→Vì}$ `x^6 >= 0`

$\text{Mà}$ `x^6 = -1`

`⇒` $\text{Không có giá trị}$ `x` $\text{nào thỏa mãn đề bài}$

$\text{→Áp dụng:}$

`+)` `A^3+B^3 = (A+B)(A^2-AB+B^2)`

`+)` `A^3-B^3 = (A-B)(A^2+AB+B^2)`

`2)`

`(2n+3)^2-(2n-1)^2`

`= (2n+3-2n+1)(2n+3+2n-1)`

`= 4(4n-2)`

`= 8(2n-1)` `\vdots` `8`

`⇔` `(2n+3)^2-(2n-1)^2` `\vdots` `8`  `AA` `n` `in` `ZZ`

`\text{Vậy biểu thức trên chia hết cho}` `8`

`\text{→Áp dụng:}` `A^2-B^2 = (A-B)(A+B)`

Bài 1:

`a)`

`(x+1)^3+(x-2)^3-2x^2(x-1,5)=3`

`⇔x^3+3x^2+3x+1+x^3-6x^2+12x-8-2x^3+3x^2=3`

`⇔(x^3+x^3-2x^3)+(3x^2-6x^2+3x^2)+(3x+12x)+(1-8)=3`

`⇔15x-7=3`

`⇔15x=10`

`⇔x=10/15`

`⇔x=2/3`

Vậy `x=2/3`

`b)`

`(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)=-65`

`⇔(x^3+8)(x^3-8)=-65`

`⇔(x^3)^2-8^2=-65`

`⇔x^6-64=-65`

`⇔x^6=-1(` Vô lí vì `x^6>=0AAx)`

Vậy không có giá trị `x` thoả mãn.

Bài 2:

`(2n+3)^2-(2n-1)^2`

`=(2n+3+2n-1)(2n+3-2n+1)`

`=(4n+2).4`

`=2.(2n+1).4`

`=8.(2n+1)`

`⇒8(2n+1)\vdots 8AAn∈ZZ`

`⇒(2n+3)^2-(2n-1)^2\vdots 8AAn∈ZZ`