=>\(\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}\)=\(\frac{\left(x+4\right)+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
=>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)
\(\Leftrightarrow\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+8\right)^2+8}{x+8}=\frac{\left(x+4\right)^2+4}{x+4}+\frac{\left(x+6\right)^2+6}{x+6}\)
\(\Leftrightarrow x+2+\frac{2}{x+2}+x+8+\frac{8}{x+8}=x+4+\frac{4}{x+4}+x+6+\frac{6}{x+6}\)
\(\Leftrightarrow\frac{2}{x+2}+\frac{8}{x+8}=\frac{4}{x+4}+\frac{6}{x+6}\)
\(\Leftrightarrow1-\frac{x}{x+2}+1-\frac{x}{x+8}=1-\frac{x}{x+4}+1-\frac{x}{x+6}\)
\(\Leftrightarrow\frac{x}{x+2}+\frac{x}{x+8}=\frac{x}{x+4}+\frac{x}{x+6}\)
\(\Leftrightarrow x\left(\frac{1}{x+2}+\frac{1}{x+8}-\frac{1}{x+4}-\frac{1}{x+6}\right)=0\)
\(\Leftrightarrow x\left(\frac{16x-80}{\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\16\left(x-5\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
cai dong thu 4 tu duoi len la \(2t\left(\frac{1}{t^2-9}-\frac{1}{t^2-1}\right)\)nhe mh ghi nham dau ''-'' thanh ''+''