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

Xét x=0 không phải là nghiệm của phương trình, chia cả tử và mẫu của mỗi phân thức ở VT cho x, ta được:

\(\frac{3}{x+1+\frac{1}{x}}+\frac{3}{x-1+\frac{1}{x}}=4\)

 Đặt \(x+\frac{1}{x}=y,\)khi đó phương trình có dạng:

\(\frac{3}{y+1}+\frac{3}{y-1}=4\Leftrightarrow\frac{6y}{y^2-1}=4\)

\(\Rightarrow4y^2-4=6y\Leftrightarrow4y^2-6y-4=0\)

\(\Leftrightarrow4y^2-8y+2y-4=0\)

\(\Leftrightarrow4y\left(y-2\right)+2\left(y-2\right)=0\Leftrightarrow\left(y-2\right)\left(4y+2\right)=0\)

\(\Leftrightarrow\left(y-2\right)\left(2y+1\right)=0\Leftrightarrow\orbr{\begin{cases}y=2\\y=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{x}=2\Leftrightarrow x^2+1=2x\Leftrightarrow x^2-2x+1=0\\x+\frac{1}{x}=-\frac{1}{2}\Leftrightarrow x^2+1=-\frac{1}{2}x\Leftrightarrow x^2+\frac{1}{2}x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}+\frac{15}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\\left(x+\frac{1}{4}\right)^2=-\frac{15}{16}\left(\times\right)\end{cases}}\)

Vậy pt có nghiệm duy nhất là x=1.

\(\frac{2}{x+\frac{1}{1+\frac{x+1}{x+2}}}=\frac{6}{3x-1}\)

\(\frac{2}{x+\frac{1}{\frac{x+2+x+1}{x+2}}}=\frac{6}{3x-1}\)

\(\frac{2}{x+\frac{1}{\frac{2x+3}{x+2}}}=\frac{6}{3x-1}\)

\(\frac{2}{x+\frac{x+2}{2x+3}}=\frac{6}{3x-1}\)

\(\frac{2}{\frac{2x+3+x+2}{2x+3}}=\frac{6}{3x-1}\)

\(\frac{2}{\frac{3x+5}{2x+3}}=\frac{6}{3x-1}\)

\(\frac{4x+6}{3x+5}=\frac{6}{3x-1}\)

\(\Rightarrow\left(4x+6\right)\left(3x-1\right)=6\left(3x+5\right)\)

\(\Rightarrow12x^2-4x+18x-6=18x+30\)

\(\Rightarrow12x^2-4x+18x-18x=30+6\)

\(\Rightarrow12x^2-4x-36=0\)

\(\Rightarrow3x^2-x-9=0\)

\(\Rightarrow x^2-\frac{1}{3}x-3=0\)

\(\Rightarrow x^2-2.\frac{1}{6}x+\frac{1}{36}-\frac{1}{36}-3=0\)

\(\Rightarrow\left(x-\frac{1}{6}\right)^2-\frac{109}{36}=0\)

\(\Rightarrow\left(x-\frac{1}{6}-\frac{\sqrt{109}}{6}\right)\left(x-\frac{1}{6}+\frac{\sqrt{109}}{6}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{109}}{6}\\x=\frac{1-\sqrt{109}}{6}\end{cases}}\)

làm lại nhé, chỗ kia quy đồng sai 

lần này làm theo cách khác

\(\frac{2}{x+\frac{1}{1+\frac{x+1}{x+2}}}=\frac{6}{3x-1}\)

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

\(\Rightarrow x+\frac{1}{\frac{2x+3}{x+2}}=x-\frac{1}{3}\)

\(\Rightarrow\frac{x+2}{2x+3}=\frac{-1}{3}\)

\(\Rightarrow\left(x+2\right).3=-1.\left(2x+3\right)\)

\(\Rightarrow3x+6=-2x-3\)

\(\Rightarrow3x+2x=-3-6\)

\(\Rightarrow5x=-9\)

\(\Rightarrow x=\frac{-9}{5}\)

vậy \(x=\frac{-9}{5}\)

Vừa lm xong mt bị sụp ... 

\(\frac{1}{x-1}+\frac{3}{3x+5}=\frac{2}{x+2}+\frac{1}{x+3}\)ĐKXĐ : \(x\ne1;-\frac{5}{3};-2;-3\)

\(\frac{1}{x-1}+\frac{3}{3x+5}-\frac{2}{x+2}-\frac{1}{x+3}=0\)

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

Khử mẫu và rút gọn ta đc : \(-3x^3+2x^2+45x+52=0\)

Mời cao nhân giải tiếp.

\(\frac{1}{x-1}+\frac{6}{3x+5}=\frac{2}{x+2}+\frac{1}{x+3}\)

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

\(\Leftrightarrow\frac{9x-1}{3x^2+2x-5}=\frac{3x+8}{x^2+5x+6}\)

\(\Rightarrow9x^3+44x^2+49x-6=9x^3+30x^2+x-40\)

\(\Leftrightarrow14x^2-48x+34=0\)

\(\Rightarrow14x^2-14x-34x+34=0\)

\(\Rightarrow\left(x-1\right)\left(14x-34\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\14x-34=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{17}{7}\end{cases}}}\)

Ngu nên làm dài dòng thôi

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

\(\Leftrightarrow\frac{3x.4+5\left(x-1\right)}{20}=\frac{20.5-10\left(3x-1\right)}{20}\)

\(\Rightarrow12x+5x-5=100-30x+10\)

\(\Leftrightarrow47x=115\)

\(\Leftrightarrow x=\frac{115}{47}\)

\(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)ĐKXĐ : \(x\ne\pm2\)

\(\Leftrightarrow\frac{-3\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{x-6}{3\left(x-2\right)\left(x+2\right)}\)

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

\(\Leftrightarrow\frac{-3x-6+3x^2-12-3x+6-x+6}{3\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{-7x-6+3x^2}{3\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow3x^2-7x-6=0\)

\(\Leftrightarrow3x^2-9x+2x-6=0\)

\(\Leftrightarrow3x\left(x-3\right)+2\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-2}{3}\end{cases}}\)( thỏa mãn )

Vậy....

\(a,\frac{x}{x-3}-\frac{6}{x^2-9}=\frac{1}{x+3}\) (đkxđ: x khác 3, -3)

\(\frac{x\left(x+3\right)-6}{\left(x-3\right)\left(x+3\right)}=\frac{1}{x+3}\)

\(x\left(x+3\right)-6=x-3\)

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

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

\(\Longrightarrow\left[\begin{array}{l}x=-3\left(L\right)\\ x=1\left(N\right)\end{array}\right.\)

\(b,\frac{x^2}{x-2}+\frac{x}{1-x}=\frac{4}{x^2-3x+2}\) (đkxđ: \(x\ne1,x\ne2)\)

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

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

\(x^2\left(x-1\right)-x\left(x-2\right)=4\)

\(x^3-x^2-x^2+2x=4\)

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

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

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

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

vì \(x^2+2>0\forall x\) ⇒ x - 2 = 0

⇒ x = 2 (ko thoả mãn)

vậy phương trình vô nghiệm