a) \(ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\)

 \(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+8x}=\frac{9}{52}\)

\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+10\right)\left(x+8\right)}-\frac{9}{52}=0\)

\(\Leftrightarrow\frac{104\left(x+10\right)\left(x+8\right)+260\left(x+1\right)\left(x+10\right)+104\left(x+1\right)\left(x+3\right)-9\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

Đoạn này cậu tự phân tích tử rồi rút gọn nhé :D Vì hơi dài nên viết ra đây sẽ rối, k đẹp mắt cho lắm :>

\(\Leftrightarrow\frac{-927x^2+1782x+9072-9x^4-198x^3}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x^4+22x^3+103x^2-198x-1008\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x^4-3x^3+25x^3-75x^{^2}+178x^2-534x+336x-1008\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left[x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)\right]}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x^3+25x^2+178x+336\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x^3+14x^2+11x^2+154x+24x+336\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left[x^2\left(x+14\right)+11x\left(x+14\right)+24\left(x+14\right)\right]}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x-3\right)\left(x+14\right)\left(x^2+11x+24\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)=0}\)

\(\Leftrightarrow\frac{-9\left(x+14\right)\left(x-3\right)\left(x+3\right)\left(x+8\right)}{52\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=0\)

\(\Leftrightarrow\frac{-9\left(x+14\right)\left(x-3\right)}{52\left(x+1\right)\left(x+10\right)}=0\)

\(\Leftrightarrow-9x^2-99x+378=0\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Leftrightarrow\left(x+14\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+14=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-14\\x=3\end{cases}}}\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-14;3\right\}\)

b) \(ĐKXĐ:x\ne-1\)

 \(x^2+\left(\frac{x}{x+1}\right)^2=\frac{5}{4}\)

\(\Leftrightarrow x^2+\frac{x^2}{\left(x+1\right)^2}-\frac{5}{4}=0\)

\(\Leftrightarrow\frac{4x^2\left(x^2+2x+1\right)+4x^2-5\left(x^2+2x+1\right)}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow4x^4+8x^3+4x^2+4x^2-5x^2-10x-5=0\)

\(\Leftrightarrow4x^2+8x^3+3x^2-10x-5=0\)

\(\Leftrightarrow4x^4+2x^3+6x^3+3x^2-10x-5=0\)

\(\Leftrightarrow2x^3\left(2x+1\right)+3x^2\left(2x+1\right)-5\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x^3+3x^2-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(2x^3-2x^2+5x^2-5x+5x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left[2x^2\left(x-1\right)+5x\left(x-1\right)+5\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(2x^2+5x+5\right)=0\)

\(\Leftrightarrow2x+1=0\)                                 

hoặc \(x-1=0\)                                    

hoặc \(2x^2+5x+5=0\)                   

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

hoặc \(x=1\left(tm\right)\)

hoặc \(\left(x+\frac{5}{4}\right)^2+\frac{55}{16}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là : \(S=\left\{-\frac{1}{2};1\right\}\)

c) \(x^4-3x^3+2x^2-9x+9=0\)

\(\Leftrightarrow x^4-x^3-2x^3+2x^2-9x+9=0\)

\(\Leftrightarrow x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-2x^2-9\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-3x^2\right)+\left(x^2-9\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2+x+3\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)

hoặc \(x-3=0\)

hoặc \(x^2+x+3=0\)

\(\Leftrightarrow\)\(x=1\left(tm\right)\)

hoặc \(x=3\left(tm\right)\)

hoặc \(\left(x-\frac{1}{2}\right)^2+\frac{11}{4}=0\left(ktm\right)\)

Vậy tập nghiệm của phương trình là :\(S=\left\{1;3\right\}\)

\(ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\)

\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{\left(x+3\right)-\left(x+1\right)}{\left(x+1\right)\left(x+3\right)}+\frac{\left(x+8\right)-\left(x+3\right)}{\left(x+3\right)\left(x+8\right)}+\frac{\left(x+10\right)-\left(x+8\right)}{\left(x+8\right)\left(x+10\right)}\)

\(=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)

\(\Leftrightarrow x^2+11x+10=52\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Delta=11^2+4.42=289,\sqrt{289}=17\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+17}{2}=3\\x=\frac{-11-17}{2}=-14\end{cases}}\)

Vậy pt có 2 nghiệm là 3 và -14

战哥làm dài và khó hiểu))):????

a) ĐKXĐ: x≠0

Ta có: \(\frac{9}{x}+2=-6\)

⇔\(\frac{9}{x}+2+6=0\)

⇔\(\frac{9}{x}+8=0\)

⇔\(\frac{9}{x}+\frac{8x}{x}=0\)

⇔9+8x=0

⇔8x=-9

hay \(x=-\frac{9}{8}\)

Vậy: \(x=-\frac{9}{8}\)

b) ĐKXĐ: x≠0;x≠-1;x≠-3

Ta có: \(\frac{7}{x+1}+\frac{-18x}{x\left(x^2+4x+3\right)}=\frac{-4}{x+3}\)

⇔\(\frac{7}{x+1}+\frac{-18x}{x\left(x+1\right)\left(x+3\right)}-\frac{-4}{x+3}=0\)

⇔\(\frac{7x\left(x+3\right)}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}+\frac{-18x}{\left(x+1\right)\cdot x\cdot\left(x+3\right)}-\frac{-4x\left(x+1\right)}{\left(x+3\right)\cdot x\cdot\left(x+1\right)}=0\)

⇔\(7x^2+21x-18x+4x\left(x+1\right)=0\)

\(\Leftrightarrow7x^2+21x-18x+4x^2+4x=0\)

⇔\(11x^2+7x=0\)

\(\Leftrightarrow x\left(11x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\11x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\11x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\frac{-7}{11}\end{matrix}\right.\)

Vậy: \(x=\frac{-7}{11}\)

c) ĐKXĐ: x≠1; x≠-3

Ta có: \(\frac{3x-1}{x-1}-1=\frac{2x+5}{x+3}+\frac{4}{x^2-2x+3}\)

⇔\(\frac{3x-1}{x-1}-1-\frac{2x+5}{x+3}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\frac{\left(3x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}-\frac{\left(2x+5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{4}{\left(x-1\right)\left(x+3\right)}=0\)

⇔\(\left(3x-1\right)\left(x+3\right)-\left(x-1\right)\left(x+3\right)-\left(2x+5\right)\left(x-1\right)-4=0\)

\(\Leftrightarrow3x^2+9x-x-3-\left(x^2+3x-x-3\right)-\left(2x^2-2x+5x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-\left(x^2+2x-3\right)-\left(2x^2+3x-5\right)-4=0\)

\(\Leftrightarrow3x^2+8x-3-x^2-2x+3-2x^2-3x+5-4=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow3x=-1\)

hay \(x=\frac{-1}{3}\)

Vậy: \(x=\frac{-1}{3}\)

ĐKXĐ: \(x\ne\left\{-10;-8;-3;-1\right\}\)

\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\)

\(\Leftrightarrow\frac{9}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\)

\(\Leftrightarrow\left(x+1\right)\left(x+10\right)=52\)

\(\Leftrightarrow x^2+11x-42=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\)

ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)

\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)

\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)

\(15-20x+6x-12=0\)

\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn 

ra nhiều thế