\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\left(x\ne1\right)\)

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

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

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

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

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

\(\Rightarrow3x=0\)

=> x=0 (tmđk)
Vậy x=0

1) Hình như đề bị sai rồi bạn.

Thông thường pt đã cho sẽ là \(\frac{2x}{x-2}-\frac{5}{x-3}=\frac{5}{x^2-5x+6}\)

Ta thấy \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)

Nên ĐKXĐ là \(\hept{\begin{cases}x\ne2\\x\ne3\end{cases}}\)

pt đã cho \(\Leftrightarrow\frac{2x\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{5\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow\frac{2x^2-6x-5x+10}{\left(x-2\right)\left(x-3\right)}=\frac{5}{\left(x-2\right)\left(x-3\right)}\)\(\Rightarrow2x^2-11x+5=0\)(*)

Ta có \(\Delta=\left(-11\right)^2-4.2.5=81>0\)nên pt (*) có 2 nghiệm phân biệt:

\(\orbr{\begin{cases}x_1=\frac{-\left(-11\right)+\sqrt{81}}{2.2}=5\left(nhận\right)\\x_2=\frac{-\left(-11\right)-\sqrt{81}}{2.2}=\frac{1}{2}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{1}{2};5\right\}\)

2) Nhận thấy \(3x^2-27=3\left(x^2-9\right)=3\left(x-3\right)\left(x+3\right)\)nên ĐKXĐ ở đây là \(x\ne\pm3\)

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

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

\(\Leftrightarrow\frac{1-3x-9}{3x^2-27}=\frac{1}{4}\)\(\Rightarrow-12x-32=3x^2-27\)\(\Leftrightarrow3x^2+12x+5=0\)(#)

Nhận thấy \(\Delta'=6^2-3.5=21>0\)

Vậy pt (#) có 2 nghiệm phân biệt \(\orbr{\begin{cases}x_1=\frac{-12+\sqrt{21}}{3}\left(nhận\right)\\x_2=\frac{-12-\sqrt{21}}{3}\left(nhận\right)\end{cases}}\)

Vậy pt đã cho có tập nghiệm \(S=\left\{\frac{-12\pm\sqrt{21}}{3}\right\}\)

Mình sẽ k cho bạn nào nhanh nhất nhé <3

\(\frac{1}{x-3}=a,\frac{1}{y-4}=b\)

\(hpt\Leftrightarrow\hept{\begin{cases}a+b=\frac{5}{3}\\4a-3b=\frac{3}{2}\end{cases}\Rightarrow\hept{\begin{cases}a=\frac{13}{14}\\b=\frac{31}{42}\end{cases}\Rightarrow}}\hept{\begin{cases}x=\frac{53}{13}\\y=\frac{166}{31}\end{cases}}\)

\(\Leftrightarrow\frac{6x^2+3}{24}-\frac{10x-4}{24}=\frac{6x^2-6}{24}-\frac{4x-12}{24}\)

\(\Leftrightarrow\frac{6x^2+3-10x+4}{24}=\frac{6x^2-6-4x+12}{24}\)

\(\Leftrightarrow6x^2-10x+7=6x^2-4x+6\)

\(\Leftrightarrow-6x+1=0\)

\(\Rightarrow-6x=-1\)

\(\Leftrightarrow x=\frac{1}{6}\)

Vậy ...

ĐKXĐ : \(x\ne2,x\ne4\)

Phương trình ban đầu tương đương :

\(\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{x^2-6x+8}=0\)

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

\(\Rightarrow x^2-5x+4+x^2+x-6+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Rightarrow x=0\) ( Do x = 2 không thỏa mãn ĐKXĐ )

Vậy pt đã cho có tập nghiệm \(S=\left\{0\right\}\)

\(ĐKXĐ:x\ne2;x\ne4\)

\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)

\(\Rightarrow\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{x^2-6x+8}\)

\(\Rightarrow\frac{\left(x^2-5x+4\right)+\left(x^2+x-6\right)}{x^2-6x+8}=\frac{-2}{x^2-6x+8}\)

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

\(\Rightarrow2x^2-4x-2=-2\)

\(\Rightarrow2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)

Vậy pt có 1 nghiệm duy nhất là 0

\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)

\(\Leftrightarrow\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{x^2-6x+8}=0\)

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

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

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

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

\(\Leftrightarrow2x^2+6x=0\)

\(\Leftrightarrow2x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

Vậy x=0; x=-3

ĐKXĐ

(x+1)(x+3)\(\ne\)0

<=>x+1\(\ne\)0 và x+3\(\ne\)0

<=>x\(\ne\)-1 và x\(\ne\)-3

Phương trình : \(\frac{x}{2\left(x+3\right)}+\frac{x}{2x+2}=\frac{4x}{\left(x+1\right)\left(x+3\right)}\)

<=>\(\frac{x}{2\left(x+3\right)}+\frac{x}{2\left(x+1\right)}=\frac{4x}{\left(x+1\right)\left(x+3\right)}\)

<=>\(\frac{x+1}{2\left(x+1\right)\left(x+3\right)}+\frac{x+3}{2\left(x+1\right)\left(x+3\right)}=\frac{8x}{2\left(x+1\right)\left(x+3\right)}\)

=>x+1+x+3=8x

<=>x+x-8x=-1-3

<=>-6x=-4

<=>x=2/3(thỏa ĐKXĐ)

Vậy S={2/3}