\(\Leftrightarrow1+\dfrac{2}{x+2}+1+\dfrac{8}{x+8}=1+\dfrac{4}{x+4}+1+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{1}{x+2}+\dfrac{4}{x+8}=\dfrac{2}{x+4}+\dfrac{3}{x+6}\)

\(\Leftrightarrow\dfrac{4}{x+8}-\dfrac{3}{x+6}=\dfrac{2}{x+4}-\dfrac{1}{x+2}\)

\(\Leftrightarrow\dfrac{4x+24-3\left(x+8\right)}{\left(x+8\right)\left(x+6\right)}=\dfrac{2x+4-\left(x+4\right)}{\left(x+4\right)\left(x+2\right)}\)

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

x=0 là nghiệm

x khác 0

\(\left\{{}\begin{matrix}x\ne\left\{-8;-6;-4;-2\right\}\\\left(x+4\right)\left(x+2\right)=\left(x+8\right)\left(x+6\right)\end{matrix}\right.\)<=>x^2 +6x+8 =x^2 +14x+48

-40 =8x=> x =-5 nhận

x={-5;0}

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

=>2x+10+\(\frac{2}{x+2}+\frac{8}{x+8}\)=2x+10+\(\frac{4}{x+4}+\frac{6}{x+6}\)

=>-x\(\left(\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}+\frac{1}{x+8}\right)\)=0

=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+2}-.....+\frac{1}{x+8}=0\end{cases}}\)

Voi \(\frac{1}{x+2}-....\)=0 ta co

Dat x+5=t

=>\(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}\)=0

=> \(2t\left(\frac{1}{t^2-1}+\frac{1}{t^2-9}\right)=0\)

=>t=0

=>x=-5

Vay phuong trinh co nghiem x=0;-5

toán lớp 8 mà đi giải phương trình hả má

ĐKXĐ:\(x\ne-2;-4;-6;-8\)

\(\frac{x^2+4x+6}{x+2}+\frac{x^2+16+72}{x+8}=\frac{x^2+8x+20}{x+4}+\frac{x^2+12x+42}{x+6}\)

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

\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)

\(\frac{10+32}{\left(x+2\right)\left(x+8\right)}=\frac{10x+48}{\left(x+4\right)\left(x+6\right)}\)

\(\Leftrightarrow\frac{5x+16}{\left(x+2\right)\left(x+8\right)}=\frac{5x+24}{\left(x+4\right)\left(x+6\right)}\)

\(\Leftrightarrow\left(5x+16\right)\left(x+4\right)\left(x+6\right)=\left(5x+24\right)\left(x+2\right)\left(x+8\right)\)

\(\Leftrightarrow x^2+5x=0\)(bạn tự biến đổi nhé)

\(\Leftrightarrow x=0;-5\)(tm ĐKXĐ)

Vậy phương trình có nghiệm 0;-5 (mình làm hơi tắt bn thông cảm nha)

=> \(\frac{(x+2)^2+2}{x+2}+\frac{(x+8)^2+8}{x+8}=\frac{(x+4)+4}{x+4}+\frac{(x+6)^2+6}{x+6}\)

=> 2x + 10 + \(\frac{2}{x+2}+\frac{8}{x+8}=2x+10+\frac{4}{x+4}+\frac{6}{x+6}\)

=>-x \((\frac{1}{x+2}-\frac{1}{x+4}-\frac{1}{x+6}-\frac{1}{x+8})=0\)

                              \(x=0\)

\(=>\orbr{\frac{1}{x+2}}-.....+\frac{1}{x+8}=0\)

Với \(\frac{1}{x+2}-...=0\). Ta có :

Đặt x + 5 = t

=> \(\frac{1}{t-3}-\frac{1}{t-1}-\frac{1}{t+1}+\frac{1}{t+3}=0\)

\(=>2t(\frac{1}{t^2-1}+\frac{1}{t^2-9})=0\)

=> t = 0

=> x = -5

Vậy phương trình có nghiệm x= 0 ; - 5

\(\dfrac{x^2+4x+6}{x+2}+\dfrac{x^2+16x+72}{x+8}=\dfrac{x^2+8x+20}{x+4}+\dfrac{x^2+12x+42}{x+6}\left(đkxđ:x\ne-2;-8;-4;-6\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow x+2+\dfrac{2}{x+2}+x+8+\dfrac{8}{x+8}=x+4+\dfrac{4}{x+4}+x+6+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

\(\Leftrightarrow\dfrac{2}{x+2}-1+\dfrac{8}{x+8}-1=\dfrac{4}{x+4}-1+\dfrac{6}{x+6}-1\)

\(\Leftrightarrow\dfrac{-x}{x+2}+\dfrac{-x}{x+8}=\dfrac{-x}{x+4}+\dfrac{-x}{x+6}\)

\(\Leftrightarrow\left(-x\right)\left(\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\\dfrac{1}{x+2}+\dfrac{1}{x+8}-\dfrac{1}{x+4}-\dfrac{1}{x+6}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

ĐKXĐ: x\(\ne\) -2; x\(\ne\) -4; x\(\ne\) -6; x\(\ne\) -8;

\(\Leftrightarrow\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+8\right)^2+8}{x+8}=\) \(\dfrac{\left(x+4\right)^2+4}{x+4}+\dfrac{\left(x+6\right)^2+6}{x+6}\)

\(\Leftrightarrow\left(x+2+\dfrac{2}{x+2}\right)+\left(x+8+\dfrac{8}{x+8}\right)=\)

\(\left(x+4+\dfrac{4}{x+4}\right)+\left(x+6+\dfrac{6}{x+6}\right)\)

\(\Leftrightarrow\dfrac{2}{x+2}+\dfrac{8}{x+8}=\dfrac{4}{x+4}+\dfrac{6}{x+6}\)

=> 2.(x+4)(x+8)(x+6) + 8(x+2)(x+4)(x+6)=4(x+2)(x+6)(x+8)

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

<=>(2x+8)(x² + 14x+64) + (8x+48)(x²+6x+8) - (4x+8)(x² + 14x+64)

-(6x+48)(x²+6x+8)

<=> (x² + 14x+64)(2x+8 -4x -8) + (x²+6x+8)(8x+48+6x-48)=0

<=> -2x(x² + 14x+64)+ 2x(x²+6x+8) = 0

<=> -2x³ -28x² -128x+ 2x³ +12x² +16x = 0

<=> -16x² - 112x = 0

<=> -x(16x+112) = 0

<=>\(\left[{}\begin{matrix}x=0\\16x+112=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\left(tmđk\right)\\x=7\left(tmđk\right)\end{matrix}\right.\)

vậy S={0;7}

sửa bài:

<=>﴾2x+8﴿﴾x² + 14x+48﴿ + ﴾8x+48﴿﴾x² +6x+8﴿ ‐ ﴾4x+8﴿﴾x² + 14x+48﴿

‐﴾6x+48﴿﴾x² +6x+8﴿
<=> ﴾x² + 14x+48﴿﴾2x+8 ‐4x ‐8﴿ + ﴾x² +6x+8﴿﴾8x+48+6x‐48﴿=0
<=> ‐2x﴾x² + 14x+48﴿+ 2x﴾x² +6x+8﴿ = 0
<=> ‐2x³ ‐28x² ‐96x+ 2x³ +12x² +16x = 0
<=> ‐16x² ‐ 80x = 0
<=> ‐x﴾16x+80﴿ = 0

<=>\(\left[{}\begin{matrix}x=0\\16x+80=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

vậy : S={0;-5}

x= 2125500 và x = 0 là nghiệm của phương trình