Đặt đúng môn học Toán bạn nhé. -

(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}

Ta có: (x+2)(x+4)(x+6)(x+8)+16

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

\(=\left[x^2+10x+16\right]\left[x^2+10x+24\right]+16\) (1)

Đặt \(x^2+10x+16=t\), khi đó (1) trở thành:

\(t\left(t+8\right)+16=t^2+8t+16=\left(t+4\right)^2\)

Thay \(x^2+10x+16=t\), ta có: \(\left(x^2+10x+16+4\right)^2=\left(x^2+10x+20\right)^2\)

Có gì đó sai sai á nhờ :vv?

( 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 + 20 

(*) <=> ( t - 4 )( t + 4 ) + 16

      = t² - 16 + 16

      = t² = ( x² + 10x + 20 )²

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

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

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

Đặt \(x^2+10x+20=t\)

\(\Rightarrow A=\left(t-4\right)\left(t+4\right)+16=t^2-16+16=t^2\)

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

x 16 + x 8 − 2 = ( x 8 ) 2 + x 8 − 2 = ( 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 )

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

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

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

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

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

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

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

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

\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right) \left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\)
\(\left(1\right)\Rightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\)
\(=\left(x^2+10x+20\right)^2\)
 

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

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

\(=\left(x^2-10x+16\right)\left(x^2-10x+24\right)+16\)(1) 

Đặt \(x^2-10x+20=t\)thay vào (1) ta được : 

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

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

\(=t^2\)Thay \(t=x^2-10x+20\)ta được :

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

\(=\left(x^2-2.5.x+25-25+20\right)^2\)

\(=\left[\left(x-5\right)^2-5\right]^2\)

\(=\left(x-5-\sqrt{5}\right)^2\left(x-5+\sqrt{5}\right)^2\)

ta có

\(5x=-3y=4z\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)

\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)