Bài `1`:

`a) (5 - x)(5 + x) - (2x - 1)^2`

`= 5^2 - x^2 - [(2x)^2 - 2. 2x + 1^2]`

`= 25 - x^2 - (4x^2 - 4x + 1)`

`= 25 - x^2 - 4x^2 + 4x - 1`

`= -5x^2 + 4x + 24`

`b) (x^4 + 2x^3 + 10x - 25) : (x^2 + 5)`

`= [(x^4 - 25) + (2x^3 + 10x)] : (x^2 + 5)`

`= {[(x^2)^2 - 5^2] + 2x(x^2 + 5)} : (x^2 + 5)`

`= [(x^2 + 5)(x^2 - 5) + 2x(x^2 + 5)] : (x^2 + 5)`

`= (x^2 + 5)(x^2 - 5 + 2x) : (x^2 + 5)`

`= x^2 - 5 + 2x`
Bài `2`:

`a) x^4 - 8x^2 - 9 = 0`

`=> (x^2)^2 - 2x^2. 4 + 16 - 25 = 0`

`=> (x^2)^2 - 2x^2. 4 + 4^2 = 25`

`=> (x^2 - 4)^2 = (+-5)^2`

`=>` \(\left\{\begin{matrix}x^2 - 4=5\\x^2 - 4 = -5\end{matrix}\right.\)

`=>` \(\left\{\begin{matrix}x^2 =9\\x^2 =-1\end{matrix}\right.\)

`=>` \(\left\{\begin{matrix}x=\pm3\\x \in\emptyset \end{matrix}\right.\)

Vậy `x = +-3`

`b) 6x^2 - (2x + 5)(3x - 2) = -12`

`=> 6x^2 - (2x. 3x - 2x. 2 + 5. 3x - 5. 2) = -12`

`=> 6x^2 - (6x^2 - 4x + 15x - 10) = -12`

`=> 6x^2 - 6x^2 + 4x - 15x + 10 = -12`
`=> -11x + 10 = -12`

`=> -11x = -22`

`=> x = 2`

Vậy `x = 2`

Đáp án:

Giải thích các bước giải:

a) $(5-x)(5+x)-(2x-1)^{2}$

= $25-x^{2}-(4x^{2}-4x+1)$

= $5x^{2}+4x+24$

b) $(x^{4}+2x^{3}+10x-25):(x^{2}+5)$

= $[(x^{2}-5)(x^{2}+5)+2x(x^{2}+5)]:(x^{2}+5)$

= $(x^{2}+5)(x^{2}-5+2x):(x^{2}+5)$

= $x^{2}+2x-5$

c) $(x+3)(x^{2}+3x-5)-x(x-2)^{2}$

= $x^{3}+6x^{2}+4x-15-x(x^{2}-4x+4)$

= $10x^{2}+8x-15$

d) $x^{4}-8x^{2}-9=0$

⇔ $x^{4}+x^{2}-9x^{2}-9=0$

⇔ $x^{2}(x^{2}+1)-9(x^{2}+1)=0$

⇔ $(x^{2}+1)(x^{2}-9)=0$

⇔ $x^{2}-9=0$

⇔ $(x-3)(x+3)=0$

⇔ $x=±3$

e) $6x^{2}-(2x+5)(3x-2)=-12$

⇔ $6x^{2}-6x^{2}-11x+10=-12$

⇔ $-11x=-22$

⇔ $x=2$