Câu c : \(x^4-3x^3+2x^2-9x+9=0\)
<=>\(x^4-x^3-2x^3+2x^2-9x+9=0\)
<=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)-9\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x^3-2x^2-9\right)=0\)
<=> \(x-1=0\) hoặc \(x^3-2x^2-9=0\)
Nếu x-1=0 <=> x=1
Nếu \(x^3-2x^2-9=0\)
<=> \(x^3-3x^2+x^2-9=0\)
<=>\(x^2\left(x-3\right)+\left(x-3\right)\left(x+3\right)=0\)
<=>\(\left(x-3\right)\left(x^2+x+3\right)=0\)
Vì \(x^2+x+3=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\) >0 nên x-3=0 <=> x=3
Vậy \(S=\left\{1;3\right\}\)

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

<=> \(4x^2\left(x^2+2x+2\right)=5\left(x^2+2x+1\right)\)
<=> \(4x^4+8x^3+8x^2=5x^2+10x+5\)
<=>\(4x^4+8x^3+3x^2-10x-5=0\)
<=>\(4x^4-4x^3+12x^3-12x^2+15x^2-15x+5x-5=0\)
<=>\(\left(x-1\right)\left(4x^3+12x^2+15x+5\right)=0\)
<=>\(\left(x-1\right)\left(2x+1\right)\left(2x^2+5x+5\right)=0\)
<=>x=1 hoặc \(x=\frac{-1}{2}\)
Phương trình \(2x^2+5x+5=0\) Vô nghiệm

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))):????

\(\Rightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+10x+8x+80}=\frac{9}{52}\)

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

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

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

\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\Rightarrow\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\Rightarrow\frac{9}{x^2+11x+10}=\frac{9}{52}\)

\(\Rightarrow x^2+11x+10=52\Rightarrow x^2+2\cdot\frac{11}{2}x+\frac{121}{4}-\frac{81}{4}=52\)

\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{289}{4}\Rightarrow x+\frac{11}{2}=\frac{17}{2}\Rightarrow x=\frac{17}{2}-\frac{11}{2}=\frac{6}{2}=3\Rightarrow x=3\)

\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)(ĐKXĐ: x khác -1;-3;-8;-10)

\(\Leftrightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+8x+10x+80}=\frac{9}{52}\)

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

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

\(\Leftrightarrow\frac{9x^2+99x+216}{x^4+22x^3+155x^2+374x+240}=\frac{9}{52}\)

\(\Rightarrow468x^2+5148x+11232=9x^4+198x^3+1395x^2+3366x+2160\)

\(\Leftrightarrow9x^4+198x^3+927x^2-1782x-9072=0\)

\(\Leftrightarrow x^4+22x^3+103x^2-198x-1008=0\)

\(\Leftrightarrow x^4-3x^3+25x^3-75x^2+178x^2-534x+336x-1008=0\)

\(\Leftrightarrow x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+25x^2+178x+336\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2+22x^2+66x+112x+336\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left[x^2\left(x+3\right)+22x\left(x+3\right)+112\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+22x+112\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+8x+14x+112\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left[x\left(x+8\right)+14\left(x+8\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+8\right)\left(x+14\right)=0\)

\(\Leftrightarrow\frac{\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}}{\orbr{\begin{cases}x+8=0\\x+14=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-3\left(\times\right)\\x=3\end{cases}}}{\orbr{\begin{cases}x=-8\left(\times\right)\\x=-14\end{cases}}}\)(Vì x=-3 và x=-8 không t/m ĐKXĐ)

Vậy tập nghiệm của pt là \(S=\left\{3;-14\right\}.\)

\(\dfrac{2}{x^2+4x+3}+\dfrac{5}{x^2+11x+24}+\dfrac{2}{x^2+18x+80}=\dfrac{9}{52}\\ ĐKXĐ:x\ne-1;x\ne-3;x\ne-8;x\ne-10\\ \Leftrightarrow\dfrac{2}{\left(x+1\right)\left(x+3\right)}+\dfrac{5}{\left(x+3\right)\left(x+8\right)}+\dfrac{2}{\left(x+8\right)\left(x+10\right)}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{1}{x+1}-\dfrac{1}{x+10}=\dfrac{9}{52}\\ \Leftrightarrow\dfrac{52\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}-\dfrac{52\left(x+1\right)}{52\left(x+1\right)\left(x+10\right)}=\dfrac{9\left(x+1\right)\left(x+10\right)}{52\left(x+1\right)\left(x+10\right)}\\ \Leftrightarrow52\left(x+10\right)-52\left(x+1\right)=9\left(x+1\right)\left(x+10\right)\\ \Leftrightarrow9\left(x^2+10x+x+10\right)=52\left(x+10-x-1\right)\\ \Leftrightarrow9\left(x^2+11x+10\right)=52\cdot9\\ \Leftrightarrow x^2+11x+10=52\\ \Leftrightarrow x^2+14x-3x-42=0\\ \Leftrightarrow x\left(x+14\right)-3\left(x+14\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+14\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-14\end{matrix}\right.\left(T/m\right)\)

Vậy.............

\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)

\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Rightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{3}{54}\)

\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)

\(\Leftrightarrow x^2+11x+28=54\)

\(\Leftrightarrow x^2+11x-26=0\)

Ta có \(\Delta=11^2+4.26=225,\sqrt{\Delta}=15\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-11+15}{2}=2\\x=\frac{-11-15}{2}=-13\end{cases}}\)

Vậy tập nghiệm S =  {2;-13}

ai giúp mình câu (a) với ạ

ĐKXĐ: \(x\ne\pm\frac{3}{2}\)

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

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2x-3\left(vn\right)\\2x+3=4\left(2x-3\right)\Rightarrow x=\frac{5}{2}\end{matrix}\right.\)

\(ĐKXĐ:x\ne-3;x\ne2;x\ne-1;x\ne\frac{1}{2}\)

Xét\(VT=\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}\)

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

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

\(=\frac{3x+9}{\left(x+3\right)\left(x-2\right)\left(x+1\right)}=\frac{3}{\left(x-2\right)\left(x+1\right)}\)

\(pt\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{3}{4x-2}\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=4x-2\)

\(\Leftrightarrow x^2-x-2=4x-2\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)(tm)

Vậy tập nghiệm của phương trình là {0;5}

ĐKXĐ: \(x\ne-3,2,-1\)

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

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

\(\Leftrightarrow10\left(x+1\right)\left(2x-1\right)-4\left(x-2\right)\left(2x-1\right)=3\left(x-2\right)\left(x+3\right)\left(x+1\right)\)

\(\Leftrightarrow12x^2+30x-18=3x^2+6x^2-15x-18\)

\(\Leftrightarrow12x^2+30x=3x^3+6x^2-15\)

\(\Leftrightarrow12x^2+30x-3x^3-6x^2+15x=0\)

\(\Leftrightarrow6x^2+45x-3x^2=0\)

\(\Leftrightarrow3x\left(2x+15-x^2\right)=0\)

\(\Leftrightarrow-x\left(x^2-2x-15\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}-x=0\\x-5=0\\x+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(tm\right)\\x=5\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)

Vậy: tập nghiệm của phương trình là: S = {0, 5}