a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

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

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

1)

<=> \(x^2-3x=0\)

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

x= 0 

x = 3

2) <=> \(x\left(x-3\right)=4\)

=> \(x=\dfrac{4}{x}+3\)

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

\(\Leftrightarrow x^2-3x-4=0\)

\(\Delta=b^2-4ac=\left(-3\right)^2-4\left(-4\right)=25>0\)

\(\Rightarrow\)Pt có 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+5}{2}=4\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-3-5}{2}=-1\end{matrix}\right.\)

Vậy \(S=\left\{4;-1\right\}\)

\(3,x^4-5x^2+6=0\)

Đặt \(t=x^2\left(t\ge0\right)\)

Pt trở thành

\(t^2-5t+6=0\)

\(\Delta=b^2-4ac=\left(-5\right)^2-4.6=1>0\)

\(\Rightarrow\)Pt ó 2 nghiệm pb

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2}=3\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-5-1}{2}-3\end{matrix}\right.\)

\(\Rightarrow t=x^2\Leftrightarrow t=\pm\sqrt{3}\)

Vậy \(S=\left\{\pm\sqrt{3}\right\}\)

\(4,x^3=9x\)

\(\Leftrightarrow x^3-9x=0\)

\(\Leftrightarrow x\left(x^2-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm3\end{matrix}\right.\)

Vậy \(S=\left\{0;\pm3\right\}\)

\(5,\left(x+2\right)\left(x-3\right)=x^2-4\)

\(\Leftrightarrow x^2-3x+2x-6-x^2+4=0\)

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

\(\Leftrightarrow-x=2\)

\(\Leftrightarrow x=-2\)

Vậy \(S=\left\{-2\right\}\)

giải pt sau

g) 11+8x-3=5x-3+x

\(\Leftrightarrow\) 8x + 8 = 6x - 3

<=> 8x-6x = -3 - 8

<=> 2x = -11

=> x=-\(\dfrac{11}{2}\)

Vậy tập nghiệm của PT là : S={\(-\dfrac{11}{2}\)}

h)4-2x+15=9x+4-2x

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

<=> -2x - 7x = 4 - 19

<=> -9x = -15

=> x=\(\dfrac{15}{9}=\dfrac{5}{3}\)

Vậy tập nghiệm của pt là : S={\(\dfrac{5}{3}\)}

g)\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=\dfrac{5}{3}+2x\)

<=> \(\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{5.2+6.2x}{6}\)

<=> 9x + 6 - 3x + 1 = 10 + 12x

<=> 6x + 7 = 10 + 12x

<=> 6x -12x = 10-7

<=> -6x = 3

=> x= \(-\dfrac{1}{2}\)

Vậy tập nghiệm của PT là : S={\(-\dfrac{1}{2}\)}

\(h,\dfrac{x+4}{5}-x+4=\dfrac{4x+2}{5}-5\)

<=> \(\dfrac{x+4-5\left(x+4\right)}{5}=\dfrac{4x+2-5.5}{5}\)

<=> x + 4 - 5x - 20 = 4x + 2 - 25

<=> x - 5x - 4x = 2-25-4+20

<=> -8x = -7

=> x= \(\dfrac{7}{8}\)

Vậy tập nghiệm của PT là S={\(\dfrac{7}{8}\)}

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

<=> \(\dfrac{21\left(4x+3\right)}{105}\)-\(\dfrac{15\left(6x-2\right)}{105}\)=\(\dfrac{35\left(5x+4\right)+3.105}{105}\)

<=> 84x + 63 - 90x + 30 = 175x + 140 + 315

<=> 84x - 90x - 175x = 140 + 315 - 63 - 30

<=> -181x = 362

=> x = -2

Vậy tập nghiệm của PT là : S={-2}

K) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)

<=> \(\dfrac{5\left(5x+2\right)}{30}-\dfrac{10\left(8x-1\right)}{30}=\dfrac{6\left(4x+2\right)-150}{30}\)

<=> 25x + 10 - 80x - 10 = 24x + 12 - 150

<=> -55x = 24x - 138

<=> -55x - 24x = -138

=> -79x = -138

=> x=\(\dfrac{138}{79}\)

Vậy tập nghiệm của PT là S={\(\dfrac{138}{79}\)}

m) \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

<=> \(\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

<=> 6x - 3 - 5x + 10 = x+7

<=> x + 7 = x+7

<=> 0x = 0

=> PT vô nghiệm

Vậy S=\(\varnothing\)

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

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

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

<=> \(\dfrac{13}{12}x=\dfrac{13}{12}\)

=> x= 1

Vậy S={1}

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

<=> \(\dfrac{2x-2x+1}{6}=\dfrac{x-36}{6}\)

<=> 2x -2x + 1= x-36

<=> 2x-2x-x = -37

=> x = 37

Vậy S={37}

q) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)

<=> \(\dfrac{4\left(2+x\right)-20.0,5x}{20}=\dfrac{5\left(1-2x\right)+20.0,25}{20}\)

<=> 8 + 4x - 10x = 5 - 10x + 5

<=> 4x-10x + 10x = 5+5-8

<=> 4x = 2

=> x= \(\dfrac{1}{2}\)

Vậy S={\(\dfrac{1}{2}\)}

g) \(11+8x-3=5x-3+x\)

\(\Leftrightarrow8+8x=6x-3\)

\(\Leftrightarrow8x-6x=-3-8\)

\(\Leftrightarrow2x=-11\)

\(\Leftrightarrow x=-\dfrac{11}{2}\)

h, \(4-2x+15=9x+4-2x\)

\(\Leftrightarrow-2x-9x+2x=4-4-15\)

\(\Leftrightarrow-9x=-15\)

\(\Leftrightarrow x=\dfrac{-15}{-9}=\dfrac{5}{3}\)

g) 11+8x-3=5x-3+x

=> 8x -5x -x = -3 -11+3

<=> 2x = -11

<=> x = \(\dfrac{-11}{2}\)

h)4-2x+15=9x+4-2x

=> -2x -9x +2x = 4-4-15

<=> -9x = -15

<=> x = \(\dfrac{5}{3}\)

ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)

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

Suy ra: \(x^2-3x+2-x^2-2x-2+5x=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

a) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30x}{30}+\dfrac{120}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow-24x+144=-5x+30\)

\(\Leftrightarrow-24x+5x=30-144\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: S={6}

b) Ta có: \(\dfrac{4-5x}{6}=\dfrac{2\left(-x+1\right)}{2}\)

\(\Leftrightarrow2\cdot\left(4-5x\right)=12\left(-x+1\right)\)

\(\Leftrightarrow2-10x=-12x+12\)

\(\Leftrightarrow2-10x+12x-12=0\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow2x=10\)

hay x=5

Vậy: S={5}

c) Ta có: \(\dfrac{-\left(x-3\right)}{2}-2=\dfrac{5\left(x+2\right)}{4}\)

\(\Leftrightarrow\dfrac{2\left(3-x\right)}{4}-\dfrac{8}{4}=\dfrac{5\left(x+2\right)}{4}\)

\(\Leftrightarrow6-2x-8=5x+10\)

\(\Leftrightarrow-2x+2-5x-10=0\)

\(\Leftrightarrow-7x-8=0\)

\(\Leftrightarrow-7x=8\)

hay \(x=-\dfrac{8}{7}\)

Vậy: \(S=\left\{-\dfrac{8}{7}\right\}\)

d) Ta có: \(\dfrac{7-3x}{2}-\dfrac{5+x}{5}=1\)

\(\Leftrightarrow\dfrac{5\left(7-3x\right)}{10}-\dfrac{2\left(x+5\right)}{10}=\dfrac{10}{10}\)

\(\Leftrightarrow35-15x-2x-10-10=0\)

\(\Leftrightarrow-17x+15=0\)

\(\Leftrightarrow-17x=-15\)

hay \(x=\dfrac{15}{17}\)

Vậy: \(S=\left\{\dfrac{15}{17}\right\}\)

a) Ta có: ⇔6(x+4)30−30x30+12030=10x30−15(x−2)30⇔6(x+4)30−30x30+12030=10x30−15(x−2)30

hay x=6

Vậy: S={6}

b) Ta có: 

Vậy: 7−3x2−5+x5=17−3x2−5+x5=1

Vậy: x+45−x+4=x3−x−22x+45−x+4=x3−x−22

hay x=5

Vậy: S={5}

c) Ta có: ⇔2(3−x)4−84=5(x+2)4⇔2(3−x)4−84=5(x+2)4

hay S={−87}S={−87}

d) Ta có: ⇔5(7−3x)10−2(x+5)10=1010⇔5(7−3x)10−2(x+5)10=1010

hay S={1517}

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

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

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

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

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

1,\(3x-1=0\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)

2,\(2-x=3x+1\Leftrightarrow2-1=3x+x\rightarrow1=4x\Rightarrow x=-\dfrac{1}{4}\)

3,\(2\left(x-2\right)-1=5x\Leftrightarrow2x-4-1=5x\Leftrightarrow2x-5x=4+1\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)

4,\(\dfrac{x}{3}-\dfrac{x}{5}=4\Leftrightarrow\dfrac{5x}{15}-\dfrac{3x}{15}=\dfrac{60}{15}\Rightarrow5x-3x=60\Rightarrow2x=60\Rightarrow x=\dfrac{60}{2}=30\)

5,\(\dfrac{x-1}{4}+\dfrac{2x+1}{6}=\dfrac{3}{2}\Leftrightarrow\dfrac{3\left(x-1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{18}{12}\)

\(3\left(x-1\right)+2\left(2x+1\right)=18\Leftrightarrow3x-3+4x+2=18\Leftrightarrow3x+4x=3-2+18\Rightarrow7x=19\Rightarrow x=\dfrac{19}{2}\)

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

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

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

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

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

\(\Leftrightarrow-x=-44\)

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

\(2,\left(đk:x\ne4\right)\)

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

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

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

\(\Leftrightarrow-x=13\)

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