\(\frac{12}{x-1}-\frac{8}{x+1}=1\left(ĐKXĐ:x\ne\pm1\right)\)

\(\Leftrightarrow\frac{12\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{8\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\) \(\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow\left(12x+12\right)-\left(8x-8\right)=x^2-1\)

\(\Leftrightarrow12x+12-8x+8=x^2-1\)

\(\Leftrightarrow12x+12-8x+8-x^2+1=0\)

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

\(\Leftrightarrow x^2-4x-21=0\)

\(\Leftrightarrow\left(x^2-4x+4\right)-25=0\)

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

\(\Leftrightarrow\left(x-7\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{7;-3\right\}\)

a thiếu

chỗ x phải có chữ thỏa mãn nữa nha

sorry sora cưng

\(\frac{x^3+7x^2+6x-30}{x^3-1}=\frac{x^2-x+16}{x^2+x+1}\) \(\left(ĐKXĐ:x\ne1\right)\)

\(\Leftrightarrow\frac{x^3+7x^2+6x-30}{\left(x-1\right)\left(x^2+x+1\right)}=\) \(\frac{\left(x^2-x+16\right)\left(x-1\right)}{\left(x^2+x+1\right)\left(x-1\right)}\)

\(\Rightarrow x^3+7x^2+6x-30=x^3-2x^2+17x-16\)

\(\Leftrightarrow9x^2-11x-14=0\)

\(\Leftrightarrow\left(9x^2-11x+\frac{121}{36}\right)-\frac{625}{36}=0\)

\(\Leftrightarrow\left(3x-\frac{11}{6}\right)^2-\left(\frac{25}{6}\right)^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=\frac{-7}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-\frac{7}{9}\left(tm\right)\end{cases}}\)

Vậy phương trình có tập nghiệm  \(S=\left\{2;\frac{-7}{9}\right\}\)

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

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

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

\(\Rightarrow\int^{\frac{x-1}{1}=0}_{\frac{x+1}{1}=0}\Rightarrow x=0\)

=>3x=5

\(\Rightarrow x=\frac{3}{5}\)

vậy \(x=\frac{3}{5}\) hoặc 0 

b)x = -(20309916*i+23555105)/9277755;

x = -(985155752*i-35635815)/916564140;

x = (985155752*i+35635815)/916564140;

x = (20309916*i-23555105)/9277755;

c)\(\Leftrightarrow\frac{x+2}{x-1}=\frac{1}{1}\Rightarrow\left(x+2\right)1=\left(x-1\right)1\)

vì \(\left(x+2\right)1\ne\left(x-1\right)1\)

=>x vô nghiệm hoặc đề sai

x=1+2+3=6 hê

cai gi the?????????

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

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

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

\(\Leftrightarrow\left(2x-5\right).\left(\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x^2-5x+4=x^2-5x+6\left(loai\right)\end{cases}}}\)

Vậy..

a,\(\frac{2x-5}{3}-\frac{3x-1}{2}< \frac{3-x}{5}-\frac{2x-1}{4}\)

\(\Leftrightarrow\frac{\left(2x-5\right)20}{60}-\frac{\left(3x-1\right)30}{60}< \frac{\left(3-x\right)12}{60}-\frac{\left(2x-1\right)15}{60}\)

\(\Leftrightarrow40x-100-90x+30< 36-12x-30x+15\)

\(\Leftrightarrow40x-90x+12x+30x< 36+15+100-30\)

\(\Leftrightarrow-8x< 121\)

\(\Leftrightarrow x>-\frac{378}{25}\)

súc vật tự đăng tự trả lời 

thằng c hó súc sinh.

đây mà là toán lớp 9 thì tao ko còn là đấng cứu thế nữa.

lớp 9 mà ngu đến phải đăng bài lớp 8 à

\(\frac{1}{x+1}+\frac{1}{x-1}=0\)  \(\left(ĐKXĐ:x\ne\pm1\right)\)

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

\(\Rightarrow x-1+x+1=0\)

\(\Leftrightarrow2x=0\)

\(\Leftrightarrow x=0\) (tm)

Vậy phương trình có tập nghiệm \(S=\left\{0\right\}\)

\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)  ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)

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

\(\Leftrightarrow x-3=10x-15\)

\(\Leftrightarrow x-10x=3-15\)

\(\Leftrightarrow-9x=-12\)

\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)

KL :....

\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\)   ĐKXĐ : \(x\ne0;2\)

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

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

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

\(\Leftrightarrow x\left(x+1\right)=0\)

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

KL ::

\(c,\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)    ĐKXĐ : \(x\ne\pm2\)

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

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

\(\Leftrightarrow0x=0\)

KL : PT vô số nghiệm