Đáp án:
Giải thích các bước giải:
Bài 4:
$a)x^2-2x-24_{}$
⇔ $x^2+4x-6x-24_{}$
⇔ $x(x+4)-6(x+4)_{}$
⇔ $(x+4)(x-6)_{}$
$b)x^2-7x+12_{}$
⇔ $x^2-3x-4x-12_{}$
⇔ $x(x-3)-4(x-3)_{}$
⇔ $(x-3)(x-4)_{}$
$c)x^3+6x^2+5x_{}$
⇔ $x(x^2+6x+5)_{}$
⇔ $x(x^2+x+5x+5)_{}$
⇔ $x.[ x.(x+5)+x+5]_{}$
⇔ $x(x+5)(x+1)_{}$
$d)2x^3+16x^2+30x_{}$
⇔ $2x(x^2+8x+15)_{}$
⇔ $2x(x^2+5x+3x+15)_{}$
⇔ $2x.[ x.(x+5)+3(x+5)]_{}$
⇔ $2x(x+5)(x+3)_{}$
$e)3x^2+7x-6_{}$
⇔ $3x^2+9x-2x-6_{}$
⇔ $3x(x+3)-2(x+3) _{}$
⇔ $(x+3)(3x-2)_{}$
$f)10x^2-11x-6_{}$
⇔ $10x^2+4x-15x-6_{}$
⇔ $2x(5x+2)-3(5x+2)_{}$
⇔ $(5x+2)(2x-3)_{}$
$g)8x^2+10x-3_{}$
⇔ $8x^2+12x-2x-3_{}$
⇔ $4x(2x+3)-(2x+3)_{}$
⇔ $(2x+3)(4x-1)_{}$
$h)7x^3+33x^2-10x_{}$
⇔ $x(7x^2+33x-10)_{}$
⇔ $x(7x^2+35x-2x-10)_{}$
⇔ $x.[ 7x.(x+5)-2(x+5)]_{}$
⇔ $x(x+5).(7x-2)_{}$
$Bài2:_{}$
$a)x^4+4y^4_{}$
⇔ $(x^2)^2+(2y^2)^2_{}$
⇔ $(x^2)^2+4x^2y^2+(2y^2)^2-4x^2-y^2_{}$
⇔ $(x^2+y^2)^2-(2xy)^2_{}$
⇔ $(x^2+2y^2-2xy)(x^2+2y^2+2xy)_{}$
$b)x^4+16_{}$
⇔ $(x^2)^2+4^2_{}$
⇔ $(x^2)^2+2x^2.4+4^2-2x^2.4_{}$
⇔ $(x^2+4)^2-8x^2_{}$
⇔ $(x^2+4)^2-(2_{}$ $\sqrt{2x})^2$
⇔ $(x^2+4-2_{}$ $\sqrt{2x}).(x^2+4+2$ $\sqrt{2x})$
$c)x^5+x+1_{}$
⇔ $x^5-x^2+x^2+x+1_{}$
⇔ $x^2(x^3-1)+(x^2+x+1)_{}$
⇔ $x^2(x-1)(x^2+x+1)+(x^2+x+1)_{}$
⇔ $(x^2+x+1)(x^3-x^2+1)_{}$
$d)x^{10}+x^8+1_{}$
⇔ $(x^{10}-x)+(x^8-x^2)+(x^2+x+1)_{}$
⇔ $x(x^9-1)+x^2(x^6-1)+(x^2+x+1)_{}$
⇔ $x.[(x^3)^3-1 ]+x^2.[(x^3)^3-1 ]+(x^2+x+1)_{}$
⇔ $x(x^3-1)(x^6+x^3+1)+x^2(x^3-1)(x^3+1)+(x^2+x+1)_{}$
⇔ $x(x-1)(x^2+x+1)(x^6+x^3+1)+x^2(x-1)(x^2+x+1)(x^3+1)(x^2+x+1)_{}$
⇔ $(x^2+x+1).[ x.(x-1).(x^6+x^3+1)+x^2.(x-1).(x^3+1+1]_{}$
