Question

Answer 1

Bài 1:
a. 

\(A=(x-3)^2-2x(x^2-1)=(x^2-6x+9)-(2x^3-2x)\)

\(=-2x^3+x^2-4x+9\)

b. \(B=(x+2)(x^2-3)-(x-1)(x^2+x+1)\)

\(=(x^3+2x^2-3x-6)-(x^3-1)=2x^2-3x-5\)

 

Answer 2

Bài 2:

a. $2x^2-4x-8y^2+2=2(x^2-2x+1)-8y^2$

$=2(x+1)^2-8y^2=2[(x+1)^2-(2y)^2]$
$=2(x+1-2y)(x+1+2y)$

b.

$16+2xy-x^2-y^2$

$=16-(x^2-2xy+y^2)=4^2-(x-y)^2=(4-x+y)(4+x-y)$

c.

$x^2-4+3(x-2)^2=(x-2)(x+2)+3(x-2)^2$

$=(x-2)[(x+2)+3(x-2)]=(x-2)(4x-4)=4(x-2)(x-1)$
d.

$x^4+2x^2-15=x^4-3x^2+5x^2-15$

$=x^2(x^2-3)+5(x^2-3)$

$=(x^2-3)(x^2+5)$

Answer 3

Bài 4:

$A=x^2+xy+y^2-3x-3y$

$2A=2x^2+2xy+2y^2-6x-6y$

$=(x^2+2xy+y^2)+x^2+y^2-6x-6y$

$=(x+y)^2-4(x+y)+4+(x^2-2x+1)+(y^2-2y+1)-6$

$=(x+y-2)^2+(x-1)^2+(y-1)^2-6\geq -6$

$\Rightarrow A\geq -3$

Vậy $A_{\min}=-3$ khi $x+y-2=x-1=y-1=0$

$\Leftrightarrow x=y=1$

Answer 4

Bài 3:

a. $(x-3)^2-(x-1)(x+1)=0$

$\Leftrightarrow (x^2-6x+9)-(x^2-1)=0$

$\Leftrightarrow -6x+10=0$

$\Leftrightarrow x=\frac{5}{3}$

b.

$(x-2)^2-2x+4=0$

$\Leftrightarrow (x-2)^2-2(x-2)=0$

$\Leftrightarrow (x-2)(x-2-2)=0$

$\Leftrightarrow (x-2)(x-4)=0$

$\Leftrightarrow x-2=0$ hoặc $x-4=0$

$\Leftrightarrow x=2$ hoặc $x=4$

c.

$x^2-3x=40$

$\Leftrightarrow x^2-8x+5x-40=0$

$\Leftrightarrow x(x-8)+5(x-8)=0$

$\Leftrightarrow (x-8)(x+5)=0$

$\Leftrightarrow x-8=0$ hoặc $x+5=0$

$\Leftrightarrow x=8$ hoặc $x=-5$

d.

$x^4+6x^3-54x-81=0$

$\Leftrightarrow (x^4-3x^3)+(9x^3-27x^2)+(27x^2-81x)+(27x-81)=0$

$\Leftrightarrow x^3(x-3)+9x^2(x-3)+27x(x-3)+27(x-3)=0$

$\Leftrightarrow (x-3)(x^3+9x^2+27x+27)=0$

$\Leftrightarrow (x-3)(x+3)^3=0$

$\Leftrightarrow x-3=0$ hoặc $x+3=0$

$\Leftrightarrow x=\pm 3$