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

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

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

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

\(\Leftrightarrow-8x< 121\)

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

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

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?????????

\(ĐKXĐ:x\ne2;x\ne4\)

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

\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)-\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=\frac{16}{5}\)

\(\Rightarrow\frac{x^2-7x+12-x^2+4x-4}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow\frac{-3x+8}{x^2-6x+8}=\frac{16}{5}\)

\(\Rightarrow-3x+8=\frac{16}{5}\left(x^2-6x+8\right)\)

\(\Rightarrow-3x+8=\frac{16}{5}x^2-\frac{96}{5}x+\frac{128}{5}\)

\(\Rightarrow\frac{16}{5}x^2-\frac{81}{5}x+\frac{88}{5}=0\)

Ta có \(\Delta=\frac{81^2}{5^2}-4.\frac{16}{5}.\frac{88}{5}=\frac{929}{25},\sqrt{\Delta}=\frac{\sqrt{929}}{5}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{81+\sqrt{929}}{32}\\x=\frac{81-\sqrt{929}}{32}\end{cases}}\)

a:=>x^2-1-x=2x-1

=>x^2-x-1=2x-1

=>x^2-3x=0

=>x=0(loại) hoặc x=3(nhận)

b:=>x+2=0 hoặc 5-3x=0

=>x=-2 hoặc x=5/3

c:=>20(1-2x)+6x=9(x-5)-24

=>20-40x+6x=9x-45-24

=>-34x+20=9x-69

=>-43x=-89

=>x=89/43

d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3

=>2x^2+4x-19=-2x+7

=>2x^2+6x-26=0

=>x^2+3x-13=0

=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)

e: =>(2x-3)(2x-3-x-1)=0

=>(2x-3)(x-4)=0

=>x=4 hoặc x=3/2

d,

\(|x-\frac{1}{3}|=\frac{5}{6}\Rightarrow \left[\begin{matrix} x-\frac{1}{3}=\frac{5}{6}\\ x-\frac{1}{3}=-\frac{5}{6}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{7}{6}\\ x=\frac{-1}{2}\end{matrix}\right.\)

e,

\(\frac{3}{4}-2|2x-\frac{2}{3}|=2\)

\(\Leftrightarrow 2|2x-\frac{2}{3}|=\frac{3}{4}-2=\frac{-5}{4}\)

\(\Leftrightarrow |2x-\frac{2}{3}|=-\frac{5}{8}<0\) (vô lý vì trị tuyệt đối của 1 số luôn không âm)

Vậy không tồn tại $x$ thỏa mãn đề bài.

f, 

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

\(\Leftrightarrow 6x-3=10+6x\)

\(\Leftrightarrow 13=0\) (vô lý)

Vậy không tồn tại $x$ thỏa mãn đề bài.

a,

$0-|x+1|=5$

$|x+1|=0-5=-5<0$ (vô lý do trị tuyệt đối của một số luôn không âm)

Do đó không tồn tại $x$ thỏa mãn điều kiện đề.

b,

\(2-|\frac{3}{4}-x|=\frac{7}{12}\)

\(|\frac{3}{4}-x|=2-\frac{7}{12}=\frac{17}{12}\)

\(\Rightarrow \left[\begin{matrix} \frac{3}{4}-x=\frac{17}{12}\\ \frac{3}{4}-x=\frac{-17}{12}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-2}{3}\\ x=\frac{13}{6}\end{matrix}\right.\)

c, 

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

\(2|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{4}\)

\(|\frac{1}{2}x-\frac{1}{3}|=\frac{7}{8}\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\ \frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{29}{12}\\ x=\frac{-13}{12}\end{matrix}\right.\)