Đặt đúng môn học Toán bạn nhé. -

Ta có:

 \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)

\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)

\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)

k nha!!

\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)

\(\text{Phân tích thành nhân tử :}\)

\(\left(x^2+10x+20\right)^2\)

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)

\(=\left(x^2+10x+16\right)\left(x^2+10+24\right)+16\)

\(=\left(x^2+10x+20-4\right)\left(x^2+10x+20+4\right)+16\)(*)

Đặt \(t=x^2+10x+20\)

(*)\(=\left(t-4\right)\left(t+4\right)+16\)

\(=t^2-16+16\)

\(=t^2\)

\(=\left(x^2+10x+20\right)^2\)

\(=\left(x^2+10x+20\right)\left(x^2+10x+20\right)\)

\(x^{16}+x^8-2=x^{16}-x^8+2x^8-2=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)

\(=\left(x^8+2\right)\left(x^8-1\right)=\left(x^8+2\right)\left(x^4+1\right)\left(x^4-1\right)\)

\(=\left(x^8+2\right)\left(x^4+1\right)\left(x^2+1\right)\left(x^2-1\right)\)

\(=\left(x^8+2\right)\left(x^4+1\right)\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\)

(x+2)(x+4)(x+6)(x+8)+16 

=(x+2)(x+8)(x+4)(x+6)+16

=(x²+10x+16)(x²+10x+24)+16

đặt t=x²+10x+16 ta được:

t.(t+8)+16

=t²+8t+16

=(t+4)²

thay t=x²+10x+16 ta được:

(x²+10x+16)²

=[(x+2)(x+8)]²

=(x+2)²(x+8)²

vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)²(x+8)²

(x+2)(x+4)(x+6)(x+8)+16 

=(x+2)(x+8)(x+4)(x+6)+16

=(x²+10x+16)(x²+10x+24)+16

đặt t=x²+10x+16 ta được:

t.(t+8)+16

=t²+8t+16

=(t+4)²

thay t=x²+10x+16 ta được:

(x²+10x+16)²

=[(x+2)(x+8)]²

=(x+2)²(x+8)²

vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)²(x+8)²

mình nghĩ bạn Triều Đăng làm sai rồi...

đoạn đầu đến chôc thay t là ok r nhưng sau đó bạn sai

đáng lí sau khi thay thì ta đc

^{(x^2 +10x + 20)^2 ms đúng}

\(x^{16}+x^8-2\)

\(=\left(x^8-1\right)\left(x^8+2\right)\)

\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+2\right)\)

\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+2\right)\)

= (x^16 -1)+(x^8 -1)

= (x^8 -1)(x^8 +1)+ (x^8 -1)

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

= (x^4 -1)(x^4 +1)(x^8 +2)

= (x^2 -1)(x^2 +1)(x^4 +1)(x^8 +2)

= (x-1)(x+1)(x^2 +1)(x^4 +1)(x^8 +2)

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

x^8+7x^4+16=x⁸+8x⁴+16-x⁴

=(x⁴+4)²-x⁴

=(x⁴-x²+4)(x⁴+x²+4)

x¹⁰+x⁸+16x²+24x+16=x²(x⁸+x⁶+16+24/x+16/x²)