Đặt đúng môn học Toán bạn nhé. -

a: \(x^4+x^2+2x+6\)

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

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

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 \(t=x^2+10x+20\)

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

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

\(=t^2\)

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

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

D

a, Cách 1 : \(x^2+5x+6=x^2+2x+3x+6=\left(x+2\right)\left(x+3\right)\)

Cách 2 : \(x^2+5x+6=x^2+2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}+6\)

\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}=\left(x+2\right)\left(x+3\right)\)

b, Cách 1 : \(x^2-x-6=x^2+2x-3x-6=\left(x-3\right)\left(x+2\right)\)

Cách 2 : \(x^2-x-6=x^2-x+\frac{1}{4}-\frac{1}{4}-6=\left(x-\frac{1}{2}\right)^2-\frac{25}{4}=\left(x-3\right)\left(x+2\right)\)

c, Cách 1 : \(x^2+6x+8=x^2+4x+2x+8=\left(x+2\right)\left(x+4\right)\)

Cách 2 : \(x^2+6x+8=x^2+6x+9-1=\left(x+3\right)^2-1=\left(x+2\right)\left(x+4\right)\)

d, Cách 1 : \(x^2-2x-8=x^2+2x-4x-8=\left(x-4\right)\left(x+2\right)\)

Cách 2 : \(x^2-2x-8=x^2-2x+1-9=\left(x-1\right)^2-9=\left(x-4\right)\left(x+2\right)\)

(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}

\(x^2-6x+8\)

\(=\left(x^2-6x+9\right)-1\)

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

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

\(=\left(x-4\right)\left(x-2\right)\)

(Tíck cho mìk vs nha!)

cách 2:

x²  -6x +8 = x² -2x -4x+8= x(x-2) -4(x-2)

= (x-2)(x-4)