\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{x+16}{\left(x+2\right)\left(x+14\right)}-\frac{x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(=\frac{8}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=8\)

     \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+14}\)

\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{1}{x+2}-\frac{1}{x+14}\)

\(\Leftrightarrow\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{\left(x+14\right)-\left(x+2\right)}{\left(x+2\right)\left(x+14\right)}\)

\(\Leftrightarrow x=\left(x+14\right)-\left(x+2\right)\)

\(\Leftrightarrow x=x+14-x-2\)

\(\Leftrightarrow x=\left(x-x\right)+\left(14-2\right)\)

\(\Leftrightarrow x=0+12\)

\(\Leftrightarrow x=12\)

x=12

Áp dụng tổng đặc biệt đấy

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=12\)

=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>x=12

Ta có: \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x=12\)

Vậy \(x=12\)

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}-\frac{1}{x+8}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+16}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{\left(x+16\right)-\left(x+2\right)}{\left(x+2\right)\left(x+16\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow x+16-x-2=x\)

\(\Rightarrow x=14\)

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{x+14}{\left(x+2\right)\left(x+14\right)}-\frac{x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

\(\Rightarrow\frac{12}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=> x = 12

mình cũng ko biết giải 

ĐKXĐ:\(x\ne\left\{-2;-4;-8;-14\right\}\)

\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

\(\Leftrightarrow2\left(x+8\right)\left(x+14\right)+4\left(x+2\right)\left(x+14\right)+6\left(x+2\right)\left(x+4\right)=x\left(x+8\right)\left(x+14\right)\)

\(\Leftrightarrow2x^2+44x+224+4x^2+64x+112+6x^2+36x+48=x^3+22x^2+112x\)

\(\Leftrightarrow12x^2+144x+384=x^3+22x^2+112x\)

\(\Leftrightarrow x^3+22x^2-12x^2+112x-144x-384=0\)

\(\Leftrightarrow x^3+10x^2-32x-384=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2+16x+64\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+8\right)^2=0\)

\(\Leftrightarrow x=6\)(x=-8 loại vì x=-8 thì PT không xác định)

x=6

cần lời giải ko

Điều kiện: x+ 2 \(\ne\) 0 ; x+ 4 \(\ne\) 0; x+ 8 \(\ne\) 0 ; x + 14 \(\ne\) 0

<=> \(\frac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\frac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\frac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

<=> \(\frac{x+4}{\left(x+2\right)\left(x+4\right)}-\frac{x+2}{\left(x+2\right)\left(x+4\right)}+\frac{x+8}{\left(x+4\right)\left(x+8\right)}-\frac{x+4}{\left(x+4\right)\left(x+8\right)}+\frac{x+14}{\left(x+8\right)\left(x+14\right)}-\frac{x+8}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)<=> \(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+4\right)}\)

<=> \(\frac{12\left(x+4\right)}{\left(x+2\right)\left(x+14\right)\left(x+4\right)}=\frac{x\left(x+14\right)}{\left(x+2\right)\left(x+4\right)\left(x+14\right)}\)

<=> 12(x + 4) = x (x + 14)

<=> 12x + 48 = x² + 14 x

<=> x² + 2x - 48 = 0 

<=> x² + 8x - 6x - 48 = 0 

<=> x(x + 8) - 6 (x + 8) = 0 

<=> (x - 6)(x + 8) = 0 <=> x - 6 = 0 (do x + 8 \(\ne\) 0)

<=> x = 6 

Vậy x = 6