\(\frac{3x-3}{x^2-1}=\frac{x}{x-2}-1\)ĐKXĐ : \(x\ne\pm1;x\ne2\)

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

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

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

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

\(\Leftrightarrow x-8=0\)

\(\Leftrightarrow x=8\)( thỏa )

Vậy....

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

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

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

ĐKXĐ : \(x\ne-1,2\)

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

\(\Leftrightarrow\)\(3x-6=x^2+x-\left(x^2-2x+x-2\right)\)

\(\Leftrightarrow\)\(3x-6=x^2+x-x^2+x+2\)

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

\(\Leftrightarrow\) \(x=8\)

Vậy phương trình có nghiệm là : \(x=8\)

giúp mình nhanh với

a, \(ĐKXĐ:x\ne2\)

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

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

\(\Rightarrow1+3x-6=3-x\)

\(\Leftrightarrow1+3x-6-3+x=0\)

\(\Leftrightarrow4x-8=0\)

\(\Leftrightarrow4x=8\)

\(\Leftrightarrow x=2\left(ktm\right)\)

vậy x thuộc tập hợp rỗng

b, \(ĐKXĐ:x\ne\pm1\)

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x-1=0\Rightarrow x=1\left(ktm\right)\end{cases}}\)

vậy x = 0

c, \(ĐKXĐ:x\ne\pm\frac{1}{2}\)

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

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

\(\Leftrightarrow\frac{32x^2}{12\left(1-2x\right)\left(2x+1\right)}=\frac{-8x\left(2x+1\right)}{12\left(1-2x\right)\left(2x+1\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(2x+1\right)}\)

\(\Rightarrow32x^2=-16x^2-8x-3+6x-24x+48x\)

\(\Leftrightarrow48x^2=22x-3\)

\(\Leftrightarrow48x^2-22x+3=0\)

\(1.\frac{7x-3}{x-1}=\frac{2}{3}\)   ( \(x\ne1\))

\(\Leftrightarrow\frac{3\left(7x-1\right)}{3\left(x-1\right)}=\frac{2\left(x-1\right)}{3\left(x-1\right)}\)

\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\frac{7}{19}\)

\(2.\frac{5x-1}{3x+2}=\frac{5x-7}{3x-1}\)

\(\Leftrightarrow\frac{\left(5x-1\right)\left(3x-1\right)}{\left(3x+2\right)\left(3x-1\right)}=\frac{\left(5x-7\right)\left(3x+2\right)}{\left(3x-1\right)\left(3x+2\right)}\)

\(\Rightarrow\left(5x-1\right)\left(3x-1\right)=\left(5x-7\right)\left(3x+2\right)\)

\(\Leftrightarrow15x^2-5x-3x+1=15x^2+10x-21x-14\)

\(\Leftrightarrow15x^2-8x+1=15x^2-11x-14\)

\(\Leftrightarrow\left(15x^2-15x^2\right)+\left(-8x+11x\right)=-14-1\)

\(\Leftrightarrow3x=-15\)

\(\Leftrightarrow x=-5\)

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

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

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

\(\Leftrightarrow3x-1-3x^2+x+3\left(3x^2-x+3x-1\right)=2x^2+2x+3x+3\)

\(\Leftrightarrow3x-1-3x^2+x+9x^2-3x+9x-3=2x^2+2x+3x+3\)

\(\Leftrightarrow6x^2+10x-4=2x^2+5x+3\)

\(\Leftrightarrow\left(6x^2-2x^2\right)+\left(10x-5x\right)=7\)

\(\Leftrightarrow4x^2+5x-7=0\)

\(\Leftrightarrow\left(2x\right)^2+4x.\frac{5}{4}+\frac{16}{25}+\frac{191}{25}=0\)

\(\Leftrightarrow\left(2x+\frac{5}{4}\right)^2-\frac{191}{25}=0\)

\(\left(2x+\frac{5}{4}\right)^2>0\)

\(\Rightarrow\left(2x+\frac{5}{4}\right)^2+\frac{191}{25}>0\)

=> PT vô nghiệm 

\(4.\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\frac{\left(1-6x\right)\left(x+2\right)}{x^2-4}+\frac{\left(9x+4\right)\left(x-2\right)}{x^2-4}=\frac{2\left(3x-2\right)+1}{x^2-4}\)

\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3\left(3x-2\right)+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8=3x^2-2x+1\)

\(\Leftrightarrow3x^2-25x-6=3x^2-2x+1\)

\(\Leftrightarrow\left(3x^2-3x^2\right)+\left(-25x+2x\right)+\left(-6-1\right)=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=\frac{-7}{23}\)

\(5.\frac{3x+2}{3x-2}-\frac{6}{2+3x}=\frac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\frac{\left(3x+2\right)^2}{9x^2-4}-\frac{6\left(3x-2\right)}{9x^2-4}=\frac{9x^2}{9x^2-4}\)

\(\Rightarrow\left(3x+2\right)^2-6\left(3x-2\right)=9x^2\)

\(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)

\(\Leftrightarrow\left(9x^2-9x^2\right)+\left(12x-18x\right)+\left(4+12\right)=0\)

\(\Leftrightarrow-6x+16=0\)

\(\Leftrightarrow-6x=-16\)

\(\Leftrightarrow x=\frac{16}{6}\)

\(6.1+\frac{1}{x+2}=\frac{12}{8-x^3}\)

\(\Leftrightarrow\frac{\left(x+2\right)\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}+\frac{1\left(8-x^3\right)}{\left(x+2\right)\left(8-x^3\right)}=\frac{12\left(x+2\right)}{\left(x+2\right)\left(8-x^3\right)}\)

\(\Rightarrow\left(x+2\right)\left(8-x^3\right)+1\left(8-x^3\right)=12\left(x+2\right)\)

\(\Leftrightarrow8x+x^4+16+2x^3+8-x^3=12x+24\)

\(\Leftrightarrow x^4+\left(2x^3-x^3\right)+\left(8x-12x\right)+\left(16-24\right)=0\)

\(\Leftrightarrow x^4+x^3-4x-8=0\)

\(\Leftrightarrow\left(x^4-4x\right)+\left(x^3-8\right)=0\)

Đến đấy mk tắc r xl bạn nhé 

Mình nghĩ tại vì :

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

Xét trường hợp \(x\)nguyên dương ta có :

\(\frac{1}{x}>\frac{1}{x+2}\)và \(\frac{1}{x+1}>\frac{1}{x+3}\)

\(\Rightarrow\)\(\frac{1}{x}+\frac{1}{x+1}>\frac{1}{x+2}+\frac{1}{x+2}\)

\(\Rightarrow\)\(\left(\frac{1}{x}+\frac{1}{x+1}\right)-\left(\frac{1}{x+2}+\frac{1}{x+3}\right)>0\)

Xét trường hợp \(x\)nguyên âm ta có :

\(\frac{1}{x}< \frac{1}{x+2}\)và \(\frac{1}{x+1}< \frac{1}{x+3}\)

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

\(\Rightarrow\)\(\left(\frac{1}{x}+\frac{1}{x+1}\right)-\left(\frac{1}{x+2}+\frac{1}{x+3}\right)< 0\)

Loại trường hợp \(x=0\)vì mẫu phải khác \(0\)

Mình nghĩ vậy :))

Ta có :

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

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

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

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

Vì \(\left(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}+\frac{1}{x+3}\right)\ne0\)

\(\Rightarrow\)\(x+5=0\)

\(\Rightarrow\)\(x=-5\)

Vậy \(x=-5\)

Phùng Minh Quân bạn có thể chứng minh cái trong ngoặc khác 0 không?

\(x-\frac{2x+1}{2}-\frac{x+2}{3}>11\)

\(\Leftrightarrow\frac{6x}{6}-\frac{3.\left(2x+1\right)}{6}-\frac{2.\left(x+2\right)}{6}>11\)

\(\Leftrightarrow\frac{6x-6x-3-2x-4}{6}>11\)

\(\Leftrightarrow\frac{-2x-7}{6}>11\)

\(\Leftrightarrow-2x-7>66\)

\(\Leftrightarrow-2x>73\)

\(\Leftrightarrow x< \frac{-73}{2}\)

Chọn B.

\(\frac{x-1}{2018}+\frac{x-2}{2017}+\frac{x-3}{2016}+\frac{x-2043}{8}\)\(=0\)

\(\Leftrightarrow\)\(\frac{x-1}{2018}-1+\frac{x-2}{2017}-1+\frac{x-3}{2016}-1\)\(+\frac{x-2043}{8}+3=0\)

\(\Leftrightarrow\)\(\frac{x-1}{2018}-\frac{2018}{2018}+\frac{x-2}{2017}-\frac{2017}{2017}\)\(+\frac{x-3}{2016}-\frac{2016}{2016}+\frac{x-2043}{8}+\frac{24}{8}=0\)

\(\Leftrightarrow\)\(\frac{x-2019}{2018}+\frac{x-2019}{2017}+\frac{x-2019}{2016}\)\(+\frac{x-2019}{8}=0\)

\(\Leftrightarrow\)\(\left(x-2019\right).\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{8}\right)=0\)

\(\Leftrightarrow\)\(x-2019=0\) ( Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{8}\ne0\))

\(\Leftrightarrow\) \(x=2019\)

Vậy phương trình có nghiệm là : \(x=2019\)