a, \(\dfrac{6-x}{4x-3}=\dfrac{2}{4x-3}\)

ĐKXĐ: \(x\ne\dfrac{3}{4}\)

PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(6-x\right)\left(4x-3\right)}{4x-3}=\dfrac{2\left(4x-3\right)}{4x-3}\)

                  \(\Rightarrow6-x=2\)

                  \(\Leftrightarrow x=4\)(thỏa mãn ĐKXĐ)

b, \(\dfrac{3-x}{2x-3}+x-1=\dfrac{-4}{2x-3}\)

ĐKXĐ: \(x\ne\dfrac{3}{2}\)

PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(3-x\right)\left(2x-3\right)}{2x-3}+\left(x+1\right)\left(2x-3\right)=\dfrac{-4\left(2x-3\right)}{2x-3}\)

                  \(\Rightarrow3-x+2x-3x+2x-3=-8x+12\)

                  \(\Leftrightarrow8x=12\)

                  \(\Leftrightarrow x=\dfrac{3}{2}\)(không thỏa mãn ĐKXĐ)

Vậy \(x\in\varnothing\).

a) ĐK: \(x\ne\dfrac{3}{4}\)

PT \(\Rightarrow27x-18-4x^2=8x-6\)

\(\Leftrightarrow4x^2-19x+12=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{3}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy phương trình có nghiệm \(x=4\)

b) ĐK: \(x\ne\dfrac{3}{2}\)

PT \(\Rightarrow3-x+2x^2-5x+3=-4\) 

\(\Leftrightarrow x^2-3x+5=0\) (Vô nghiệm)

  Vậy phương trình vô nghiệm

c) ĐK: \(x\ne3\)

PT \(\Rightarrow2x^2-5x-3=2x-4\)

\(\Leftrightarrow2x^2-7x+1=0\) \(\Leftrightarrow x=\dfrac{7\pm\sqrt{41}}{4}\)

  Vậy phương trình có nghiệm \(x=\dfrac{7\pm\sqrt{41}}{4}\)

d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)

\(\Leftrightarrow8x+4-6+6x\ge12-3x\)

\(\Leftrightarrow14x+3x\ge12+2=14\)

\(\Leftrightarrow x\ge\dfrac{14}{17}\)

e: Ta có: \(\dfrac{x+1}{2}-\dfrac{2-x}{3}< \dfrac{2x-3}{4}\)

\(\Leftrightarrow6x+12+4x-8< 6x-9\)

\(\Leftrightarrow4x< -9+8-12=-13\)

hay \(x< -\dfrac{13}{4}\)

a: \(\Leftrightarrow30\left(x-3\right)-16=9\left(x-1\right)+72\)

\(\Leftrightarrow30x-90-16=9x-9+72\)

=>30x-106=9x+63

=>21x=169

hay x=169/21

b: =>4x+20=2x-3

=>2x=-23

hay x=-23/2

1: Sửa đề: 2/x+2

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

=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

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

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

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

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

=>x=0(nhận) hoặc x=3(loại)

đk : x khác -1 ; 3 

\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x\Leftrightarrow2x^2-2x-4x=0\)

\(\Leftrightarrow2x^2-6x=0\Leftrightarrow2x\left(x-3\right)=0\Leftrightarrow x=0;x=3\left(ktm\right)\)

a: =>\(\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{4x}{2\left(x-3\right)\left(x+1\right)}=\dfrac{-x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)

=>x^2-3x-4x=-x^2-x

=>x^2-7x+x^2+x=0

=>2x^2-6x=0

=>x=0(nhận) hoặc x=3(loại)

b: =>\(\dfrac{2x-3-3x-15}{x+5}>=0\)

=>\(\dfrac{-x-18}{x+5}>=0\)

=>x+18/x+5<=0

=>-18<=x<-5

\(\dfrac{x}{2x+1}-\dfrac{2x}{x^2-2x-3}=\dfrac{x}{6-2x}\) (ĐKXĐ: \(x\ne3;x\ne-1\)

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

\(\Leftrightarrow\dfrac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\dfrac{2.2x}{2\left(x-3\right)\left(x+1\right)}=-\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}\)

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

\(\Leftrightarrow x^2-7x=-x^2-x\)

\(\Leftrightarrow x^2+x^2-7x+x=0\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)

*TM: Thỏa mãn, KTM: Ko thỏa mãn

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

\(\dfrac{2x-3}{x+5}\ge3\) (ĐKXĐ: \(x\ne-5\)

\(\Leftrightarrow\dfrac{2x-3}{x+5}-3\ge0\)

\(\Leftrightarrow\dfrac{2x-3}{x+5}-\dfrac{3x+15}{x+5}\ge0\)

\(\Leftrightarrow\dfrac{2x-3-3x-15}{x+5}\ge0\)

\(\Leftrightarrow\dfrac{-x-18}{x+5}\ge0\)

\(\Leftrightarrow-18\le x\le-5\)