\(1,\left(dk:x\ne0,-1,4\right)\)

\(\Leftrightarrow\dfrac{9}{x+1}+\dfrac{2}{x-4}-\dfrac{11}{x}=0\)

\(\Leftrightarrow\dfrac{9x\left(x-4\right)+2x\left(x+1\right)-11\left(x+1\right)\left(x-4\right)}{x\left(x+1\right)\left(x-4\right)}=0\)

\(\Leftrightarrow9x^2-36x+2x^2+2x-11x^2+44x-11x+44=0\)

\(\Leftrightarrow-x=-44\)

\(\Leftrightarrow x=44\left(tm\right)\)

\(2,\left(đk:x\ne4\right)\)

\(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{2+x}{x-4}-\dfrac{3}{2\left(x-4\right)}+\dfrac{5}{6}=0\)

\(\Leftrightarrow\dfrac{14.2-6\left(2+x\right)-3.3+5\left(x-4\right)}{6\left(x-4\right)}=0\)

\(\Leftrightarrow28-12-6x-9+5x-20=0\)

\(\Leftrightarrow-x=13\)

\(\Leftrightarrow x=-13\left(tm\right)\)

\(\Leftrightarrow3\left(5x-1\right)+5\left(2x+3\right)>2\left(x-8\right)-x+1\)

=>15x-3+10x+15>2x-16-x+1

=>25x+12>x-15

=>24x>-27

hay x>-9/8

a:=>3x=15

=>x=5

b: =>8-11x<52

=>-11x<44

=>x>-4

c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)

\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)

a, đk : x khác 5;-6 

\(x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

\(\Leftrightarrow2x+61=23x+61\Leftrightarrow21x=0\Leftrightarrow x=0\)(tm) 

b, đk : x khác 1;3 

\(x^2+2x-15=x^2-1-8\Leftrightarrow2x-15=-9\Leftrightarrow x=3\left(ktmđk\right)\)

pt vô nghiệm 

a, đk : x khác 5;-6 

(tm) 

b, đk : x khác 1;3 

x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)x2+2x−15=x2−1−8⇔2x−15=−9⇔x=3(ktmđk)

pt vô nghiệm 

a: \(\Leftrightarrow\left(x+6\right)^2+\left(x-5\right)^2=2x^2+23x+61\)

\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)

=>x=0(nhận)

b: \(\Leftrightarrow\left(x+5\right)\left(x-3\right)=\left(x+1\right)\left(x-1\right)-8\)

\(\Leftrightarrow x^2+2x-15=x^2-1-8\)

=>2x-15=-9

=>2x=-6

hay x=-3(nhận)

a)Để biểu thức vô nghĩa thì \(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\Leftrightarrow x\notin\left\{-2;1\right\}\)

b) Ta có: \(\dfrac{5x-2}{12}-\dfrac{2x^2+1}{8}=\dfrac{x-3}{6}+\dfrac{1-x^2}{4}\)

\(\Leftrightarrow\dfrac{2\left(5x-2\right)}{24}-\dfrac{3\left(2x^2+1\right)}{24}=\dfrac{4\left(x-3\right)}{24}+\dfrac{6\left(1-x^2\right)}{24}\)

\(\Leftrightarrow10x-4-6x^2-3=4x-12+6-6x^2\)

\(\Leftrightarrow-6x^2+10x-7+6x^2-4x+6=0\)

\(\Leftrightarrow6x-1=0\)

\(\Leftrightarrow6x=1\)

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

Vậy: \(S=\left\{\dfrac{1}{6}\right\}\)

a. ĐKXĐ: \(x\ne2\).

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

⇔\(\dfrac{1}{x-2}+\dfrac{3x-6}{x-2}=\dfrac{3-x}{x-2}\)

⇔\(1+3x-6=3-x\)

⇔\(4x-8=0\)

⇔\(x=2\) (không thỏa mãn)

-Vậy S=∅.

b. ĐKXĐ: \(x\ne-1\)

 \(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)

⇔\(\dfrac{5x}{2\left(x+1\right)}+1=-\dfrac{6}{x+1}\)

⇔\(\dfrac{5x}{2\left(x+1\right)}+\dfrac{2\left(x+1\right)}{2\left(x+1\right)}=-\dfrac{12}{2\left(x+1\right)}\)

⇔\(5x+2\left(x+1\right)=-12\)

⇔\(5x+2x+2+12=0\)

⇔\(7x+14=0\)

⇔\(x=-2\) (thỏa mãn).

-Vậy \(S=\left\{-2\right\}\)

a, \(\Leftrightarrow\dfrac{1}{x-2}+\dfrac{3.\left(x-2\right)}{x-2}=\dfrac{3-x}{x-2}\\ \Leftrightarrow1+3x-6=3-x\)

\(\Leftrightarrow3x+x=3-1+6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=\dfrac{8}{4}=2\\ Vậy.S=\left\{2\right\}\)

b,  \(\Leftrightarrow\)\(\dfrac{5x}{2x+2}+\dfrac{2x+2}{2x+2}=\dfrac{-6.2}{2.\left(x+1\right)}\)

\(\Leftrightarrow5x+2x+2=-12\\ \Leftrightarrow7x=-12-2\\ \Leftrightarrow7x=-14\\ \Leftrightarrow x=-\dfrac{14}{7}=-2\\ Vậy.S=\left\{-2\right\}\)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

a) \(\dfrac{3}{x-7}+\dfrac{2}{x+7}=\dfrac{5}{x^2-49}\)

(ĐKXĐ: x khác 7; x khác -7)

<=>\(\dfrac{3.\left(x+7\right)}{\left(x-7\right).\left(x+7\right)}+\dfrac{2.\left(x-7\right)}{\left(x+7\right).\left(x-7\right)}=\dfrac{5}{\left(x+7\right).\left(x-7\right)}\)

=> 3x + 21 + 2x - 14 = 5

<=> 3x + 2x = 5 + 14 - 21

<=> 5x = -2

<=> x = \(\dfrac{-2}{5}\)

Vậy S = { \(\dfrac{-2}{5}\) }

b) \(\dfrac{2x-1}{3}-\dfrac{x+3}{2}>1+\dfrac{5x}{6}\)

<=> \(\dfrac{2.\left(2x-1\right)}{3.2}-\dfrac{3.\left(x+3\right)}{3.2}>\dfrac{1.6}{6}+\dfrac{5x}{6}\)

=> 4x - 2 - 3x - 9 > 6 + 5x

<=> 4x - 3x - 5x > 6 + 9 + 2

<=> -4x > 17

<=> \(\dfrac{-17}{4}\)

Vậy S = { \(\dfrac{-17}{4}\) }

g.\(\dfrac{1-3x}{6}+x-1=\dfrac{x+2}{2}\)

\(\Leftrightarrow\dfrac{\left(1-3x\right)+6\left(x-1\right)}{6}=\dfrac{3\left(x+2\right)}{6}\)

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

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

\(\Leftrightarrow-5=6\left(vô.lí\right)\)

Vậy pt vô nghiệm

h.\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)

\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)

\(\Leftrightarrow30x+15-100-6x-4=24x-8\)

\(\Leftrightarrow-89=-8\left(vô.lí\right)\)

Vậy pt vô nghiệm

i.\(\dfrac{4x+3}{5}-\dfrac{6x-2}{7}=\dfrac{5x+4}{3}+3\)

\(\Leftrightarrow\dfrac{21\left(4x+3\right)-15\left(6x-2\right)}{105}=\dfrac{35\left(5x+4\right)+215}{105}\)

\(\Leftrightarrow21\left(4x+3\right)-15\left(6x-2\right)=35\left(5x+4\right)+215\)

\(\Leftrightarrow84x+63-90x+30=175x+140+215\)

\(\Leftrightarrow-181=262\)

\(\Leftrightarrow x=-\dfrac{262}{181}\)