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

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

M = (x^2 + 2x + 8x + 16)(x^2 + 4x + 6x + 24) + 16

M = (x^2 + 10x + 16)(x^2 + 10x + 24) + 16

Đặt t = x^2 + 10x + 20

M = (t - 4)(t + 4) + 16

M = t^2 - 16 + 16 = t^2

Vậy ta có đpcm

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

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

\(M=\left(x^2+16+10x\right)\left(x^6+10x+16+8\right)+16\)

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

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

M= (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

=(x²+10x+16)(x²+10x+16+8)+16

=(x²+10x+16)²+8(x²+10x+16)+16

=(x²+10x+20)²

=>dpcm

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

=(x²+10x+16)(x²+10x+24)+16

=(x²+16+10x)(x²+10x+16+8)+16

=(x²+10x+16)²+8(x²+10x+16)+16

=(x²+10x+20)²

ĐPCM

\(\text{a, Ta có :}\) \(M=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)

\(\text{Đặt }a=x^2+10x+16\)

\(\text{Ta có: }M=a\left(a+8\right)+16=a^2+8a+16=\left(a+4\right)^2\)

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

\(\text{b, }\)\(\left|x+1\right|=\left|x\left(x+1\right)\right|\)

\(\Leftrightarrow\left|x\left(x+1\right)\right|-\left|x+1\right|=0\)

\(\Leftrightarrow\left|x\right|.\left|x+1\right|-\left|x+1\right|=0\)

\(\Rightarrow\left|x+1\right|\left(\left|x\right|-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left|x+1\right|=0\\\left|x\right|-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

a là một số bất kỳ à:

í lộn , a thay vào là 4 cho mình với

\(M=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\Leftrightarrow y\left(y+2\right)\left(y+4\right)\left(y+6\right)+16\)Y^2+6y)(y^2+6y+8)

\(M=z\left(z+8\right)+16=z^2+8z+16=\left(z+4\right)^2=K^2\) 

x thuộc Q=> y thuộc Q=> z thuộc Q=> K thuộc Q=> dpcm