Bài 1 :

\(x^2-6x+8=x^2-2x-4x+8=x\left(x-2\right)-4\left(x-2\right)=\left(x-4\right)\left(x-2\right)\)

Bài 2 :

 \(x^8+x^7+1=x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1-x^6-x^5-x^4-x^3-x^2-x\)

\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+x^2+x+1-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)

=\(\left(x^2+x+1\right)\left(x^6+x^3+1-x^4-x\right)\)

Tick đúng nha 

X^2-6+8

a: =(16x+20)^2-(10x+10)^2

=(16x+20-10x-10)(16x+20+10x+10)

=(26x+30)(6x+10)

=4(13x+15)(3x+5)

b: =(x-y+4-2x-3y+1)(x-y+4+2x+3y-1)

=(-x-4y+5)(3x+2y+3)

c: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(x^2+2x+1-x^2+2x-1)(x^2+2x+1+x^2-2x+1)

=2(x^2+1)*4x

=8x(x^2+1)

$16(4x+5)^2-25(2x+2)^2$

$=[4(4x+5)]^2-[5(2x+2)]^2$

$=[4(4x+5)-5(2x+2)][4(4x+5)+5(2x+2)]$

$=(16x+20-10x-10)(16x+20+10x+10)$

$=(6x+10)(26x+30)$

$=4(3x+5)(13x+15)$

$(x-y+4)^2-(2x+3y-1)^2$

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

$=(-x-4y+5)(3x+2y+3)$

Vậy:

$=-(x+4y-5)(3x+2y+3)$

$(x+1)^4-(x-1)^4$

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

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

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

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

$=8x(x^2+1)$

d)

Câu này không phân tích được thành nhân tử với hệ số nguyên/rational theo các hằng đẳng thức thông thường: $x^6-y^5$

\(=\left(x^2-6x+9\right)-2=\left(x-3\right)^2-\sqrt{2^2}=\left(x-3-\sqrt{2}\right)\left(x-3+\sqrt{2}\right)\)

\(=\left(x^3-6x^2+12x-8\right)+1\\ =\left(x-2\right)^3+1\\ =\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)\\ =\left(x-1\right)\left(x^2-5x+7\right)\)

ta có:( x²-2x3+3²⁾-4= (x-3)²-2²=[(x-3)-2][(x-3)+2]

a) \(x^5+x+1=x^5+x^2-x^2+x+1\)

\(=\left(x^5-x^2\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)

b) \(x^7+x^2+1=x^7+x^2-x+x+1\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x\left(x^6-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\)

\(=x\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left[x\left(x^3+1\right)\left(x-1\right)+1\right]\)

\(=\left(x^2+x+1\right)\left(x^5+x^2+1-x^4-x\right)\)

(Nếu đúng thì k cho mìk với nhé!)

Hình như ở câu phân tích đa thức đề là (3x+5)(x-2) chứ không phải (x+2)

1..

(x+1)(2-x)-(3x+5)(x+2)

=(x+1)(2-x)+(3x-5)(2-x)

=(2-x)(x+1+3x-5)

=(2-x)(4x-4)

2...

(3x-5)(2x+7)-(x+1)(6x-5)=16

=>(6x²⁺21x-10x-35)-(6x²-5x+6x-5)-16=0

=>6x²⁺21x-10x-35-6x²+5x-6x+5-16=0

=>(6x²-6x²)+(21x-10x+5x-6x)+(-35+5-16)=0

=>10x-46=0

=>10x     =46

=>    x     =4.6

 Vậy x=4.6(bạn nên đổi ra phân số thì sẽ đúng hơn)

\(x+8\sqrt{x}+7\)

\(=x+\sqrt{x}+7\sqrt{x}+7\)

\(=\sqrt{x}\left(\sqrt{x}+1\right)+7\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{x}+7\right)\)