a: \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)

=>\(9x^2+12x+4-\left(9x^2-12x+4\right)-5x-38=0\)

=>\(9x^2+7x-34-9x^2+12x-4=0\)

=>19x-38=0

=>19x=38

=>x=38/19=2

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

=>\(x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)

=>\(x^3+3x^2+12x-9=x^3+3x^2+3x+1\)

=>12x-9=3x+1

=>12x-3x=1+9

=>9x=10

=>x=10/9

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)

\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)

\(\Leftrightarrow12x-9=29x-145\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x+136=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\left(tm\right)\)

Vậy \(S=\left\{8\right\}\)

\(2,\dfrac{2x-1}{5-3x}=2\left(ĐKXĐ:x\ne\dfrac{5}{3}\right)\)

\(\Rightarrow2x-1=2\left(5-3x\right)\)

\(\Leftrightarrow2x-1=10-6x\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x-11=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{11}{8}\right\}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)

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

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

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

\(\Rightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-5-3x+2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)

\(\Rightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\)

\(2,\dfrac{2x-1}{5-3x}=2\)

\(\Leftrightarrow\dfrac{2x-1-2\left(5-3x\right)}{5-3x}=0\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\)

\(\Leftrightarrow\dfrac{4x-5-2\left(x-1-x\right)}{x-1}=0\)

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

\(\Leftrightarrow4x=3\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

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

\(\Leftrightarrow\dfrac{\left(2x+5\right)\left(x+5\right)-2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)

\(\Leftrightarrow15x=-25\)

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

a: \(x^3+8x=5x^2+4\)

=>\(x^3-5x^2+8x-4=0\)

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

=>\(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-4x+4\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2=0\)

=>\(\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

2: \(x^3+3x^2=x+6\)

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

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

=>\(x^2\cdot\left(x+2\right)+x\left(x+2\right)-3\left(x+2\right)=0\)

=>\(\left(x+2\right)\left(x^2+x-3\right)=0\)

=>\(\left[{}\begin{matrix}x+2=0\\x^2+x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1+\sqrt{13}}{2}\\x=\dfrac{-1-\sqrt{13}}{2}\end{matrix}\right.\)

3: ĐKXĐ: x>=0

\(2x+3\sqrt{x}=1\)

=>\(2x+3\sqrt{x}-1=0\)

=>\(x+\dfrac{3}{2}\sqrt{x}-\dfrac{1}{2}=0\)

=>\(\left(\sqrt{x}\right)^2+2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{17}{16}=0\)

=>\(\left(\sqrt{x}+\dfrac{3}{4}\right)^2=\dfrac{17}{16}\)

=>\(\left[{}\begin{matrix}\sqrt{x}+\dfrac{3}{4}=-\dfrac{\sqrt{17}}{4}\\\sqrt{x}+\dfrac{3}{4}=\dfrac{\sqrt{17}}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{17}-3}{4}\left(nhận\right)\\\sqrt{x}=\dfrac{-\sqrt{17}-3}{4}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{13-3\sqrt{17}}{8}\left(nhận\right)\)

4: \(x^4+4x^2+1=3x^3+3x\)

=>\(x^4-3x^3+4x^2-3x+1=0\)

=>\(x^4-x^3-2x^3+2x^2+2x^2-2x-x+1=0\)

=>\(x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-2x^2+2x-1\right)=0\)

=>\(\left(x-1\right)\left(x^3-x^2-x^2+x+x-1\right)=0\)

=>\(\left(x-1\right)^2\cdot\left(x^2-x+1\right)=0\)

=>(x-1)^2=0

=>x-1=0

=>x=1

a.

\(x^3+8x=5x^2+4\)

\(\Leftrightarrow x^3-5x^2+8x-4=0\)

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

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

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

b.

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

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{-1\pm\sqrt{13}}{2}\end{matrix}\right.\)

c.

\(2x+3\sqrt{x}+1=0\)

ĐKXĐ: \(x\ge0\)

Do \(x\ge0\Rightarrow\left\{{}\begin{matrix}2x\ge0\\3\sqrt{x}\ge0\end{matrix}\right.\)

\(\Rightarrow2x+3\sqrt{x}+1>0\)

Pt đã cho vô nghiệm

d.

\(x^4+4x^2+1=3x^3+3x\)

\(\Leftrightarrow x^4-3x^3+4x^2-3x+1=0\)

- Với \(x=0\) ko phải nghiệm

- Với \(x\ne0\) chia cả 2 vế của pt cho \(x^2\)

\(\Rightarrow x^2-3x+4-\dfrac{3}{x}+\dfrac{1}{x^2}=0\)

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

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

Đặt \(x+\dfrac{1}{x}=t\)

\(\Rightarrow t^2-3t+2=0\Rightarrow\left[{}\begin{matrix}t=1\\t=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=2\\x+\dfrac{1}{x}=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=0\left(vn\right)\\x^2-2x+1=0\end{matrix}\right.\)

\(\Rightarrow x=1\)

A 3x-4x=-9-3

    -x=-12

     x=12

B 3.2x -5x +1=5+0.2x

  3.2x-5x-0.2x=5-1

  -2x=4

 x=-2

C 1.5-x-2=-3x-0.3

  -x+3x=-0.3-1.5+2

  2x =0.2

  x=0.1

E 2/3-1/2x-1=-x+1

  -1/2x+x=1+1-2/3

  1/2x=4/3

  x=8/3

F 3t-4+13+2t+4-3t

  =3t+2t-3t-4+13+4

  =2t+13

a) 3x-6=0

3x=6 => x=2

b) (3x+2)(4x-5)=0

=> 3x+2=0 => x=-2/3

hoặc 4x-5=0 => x=5/4

câu c ,d thiếu dấu '=" để thành 1 pt rồi bạn

c) \(\dfrac{2x-5}{3}+\dfrac{x-3}{5}=\dfrac{4x+3}{15}\)

=> 10x -25 +3X-9=4X+3

=>9x=37

=>x=37/9

d) \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\) ĐK (x khác 3,-3)

=>5x+15+4x-12=x-5

=>8x=-8

=>x=-1

b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)

\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)

\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)

\(\Leftrightarrow3x\left(x+4\right)=0\)

=>x=0(nhận) hoặc x=-4(loại)

a: =>4x-2x-2-3x-2=0

=>-x-4=0

=>x=-4

b: =>x+2-2x-2+x=0

=>0x=0(luôn đúng)

d: =>3x=3

hay x=1

e: =>2x=1

hay x=1/2

f: =>4x=-4

hay x=-1

g: =>3x=-3

hay x=-1

c: =>2x+3-5-4+x=0

=>3x-6=0

=>x=2

d: =>3x=3

hay x=1

e: =>2x=1

hay x=1/2

f: =>4x=-4

hay x=-1

g: =>3x=-3

hay x=-1

\(a,4x-2\left(x+1\right)=3x+2\\ \Leftrightarrow4x-2x-2-3x-2=0\\ \Leftrightarrow-x-4=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)

Vậy pt có tập nghiệm \(S=\left\{-4\right\}\)

\(b,x+2-2\left(x+1\right)=-x\\ \Leftrightarrow x+2-2x-2+x=0\\ \Leftrightarrow0=0\)

Vậy pt có tập nghiệm \(S=R\)

\(c,2\left(x+3\right)-5=4-x\\ \Leftrightarrow2x+6-5-4+x=0\\ \Leftrightarrow3x-3=0\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)

Vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(d,3x-2=1\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)

Vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(e,2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{2}\right\}\)

\(f,4x+3=-1\\ \Leftrightarrow4x=-4\\ \Leftrightarrow x=-1\)

Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)

\(g,3x+2=-1\\ \Leftrightarrow3x=-3\\ \Leftrightarrow x=-1\)

Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)

a,4x-2(x+1)=3x+2

⇔ 4x - 2x -2 = 3x + 2

⇔ x = -4

b,x+2-2(x+1)=-x

⇔ x + 2 - 2x - 2 = -x

⇔ 0 = 0

c,2(x+3)-5=4-x

⇔ 2x + 6 - 5 = 4 - x

⇔ 3x = 3

⇔ x = 1

d,3x-2=1

⇔ 3x = 3

⇔ x = 1

e,2x-1=0

⇔ x = \(\dfrac{1}{2}\)

f,4x+3=-1

⇔ x = -1

g,3x+2=-1

⇔ x = -1