Bài 1: 

a) Ta có: \(2\left(3-4x\right)=10-\left(2x-5\right)\)

\(\Leftrightarrow6-8x-10+2x-5=0\)

\(\Leftrightarrow-6x+11=0\)

\(\Leftrightarrow-6x=-11\)

hay \(x=\dfrac{11}{6}\)

b) Ta có: \(3\left(2-4x\right)=11-\left(3x-1\right)\)

\(\Leftrightarrow6-12x-11+3x-1=0\)

\(\Leftrightarrow-9x-6=0\)

\(\Leftrightarrow-9x=6\)

hay \(x=-\dfrac{2}{3}\)

a) \(\dfrac{15-6x}{3}>5\Leftrightarrow15-6x>15\)

\(\Leftrightarrow-6x>0\Leftrightarrow x< 0\) (vì \(-6< 0\))

\(S=\left\{x|x< 0\right\}\)

b) \(\dfrac{8-11x}{4}< 13\Leftrightarrow8-11x< 52\)

\(\Leftrightarrow-11x< -44\Leftrightarrow x>4\) (vì \(-11< 0\))

\(S=\left\{x|x>4\right\}\)

c) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

\(\Leftrightarrow8x+3x+1>5x-2x+6\)

\(\Leftrightarrow8x+3x-5x+2x>6-1\)

\(\Leftrightarrow8x>5\)

\(\Leftrightarrow x>\dfrac{5}{8}\) (vì \(8>0\))

\(S=\left\{x|x>\dfrac{5}{8}\right\}\)

d) \(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)

\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)

\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>-6\)

\(\Leftrightarrow-3x>-6\)

\(\Leftrightarrow x< 2\) (vì \(-3< 0\))

\(S=\left\{x|x< 2\right\}\)

a) \(\dfrac{15-6x}{3}>5\) <=> \(15-6x>15\) <=> \(6x< 0\) <=> \(x< 0\)

b) \(\dfrac{8-11x}{4}< 13\) <=> \(8-11x< 52\) <=> \(11x>-44\)<=> \(x>-4\)

c) \(8x+3\left(x+1\right)>5x-\left(2x-6\right)\)

<=> 8x + 3x + 3 - 5x + 2x - 6 > 0

<=> 8x  > 3

<=> x > 3/8

d) 2x(6x - 1) > (3x - 2)(4x + 3)

<=> 12x² - 2x > 12x² + x - 6

<=> 12x² - 2x - 12x² - x > -6

<=> -3x > -6

<=> x < 2

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 \(x+\dfrac{1}{x}=t\)

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

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

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

\(\Rightarrow x=1\)

`|5x| = - 3x + 2`

Nếu `5x>=0<=> x>=0` thì phương trình trên trở thành :

`5x =-3x+2`

`<=> 5x +3x=2`

`<=> 8x=2`

`<=> x= 2/8=1/4` ( thỏa mãn )

Nếu `5x<0<=>x<0` thì phương trình trên trở thành :

`-5x = -3x+2`

`<=>-5x+3x=2`

`<=> 2x=2`

`<=>x=1` ( không thỏa mãn ) 

Vậy pt đã cho có nghiệm `x=1/4`

__

`6x-2<5x+3`

`<=> 6x-5x<3+2`

`<=>x<5`

Vậy bpt đã cho có tập nghiệm `x<5`

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

`<=> (5(5x+2))/30 - (10(8x-1))/30 = (6(4x+2))/30 - (5.30)/30`

`<=> 5(5x+2) - 10(8x-1) =6(4x+2) - 5.30`

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

`<=> -55x +20 = 24x-138`

`<=> -55x -24x=-138-20`

`<=>-79x=-158`

`<=> x=2`

Vậy pt có nghiệm `x=2`

`b,` \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-2\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne0\end{matrix}\right.\)

Ta có : `(x+2)/(x-2) -1/x = 2/(x(x-2))`

`<=> (x(x+2))/(x(x-2)) - (x-2)/(x(x-2))  = 2/(x(x-2))`

`=> x^2 +2x - x +2 = 2`

`<=> x^2 + x =0`

`<=>x(x+1)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)

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

Vậy pt có nghiệm `x=-1`

`c,2x^3 + 6x^2 =x^2 +3x`

`<=> 2x^3 + 6x^2 -x^2 -3x=0`

`<=> 2x^3 + 5x^2 -3x=0`

`->` Đề có sai ko ạ ?

`d,` \(\left|x-4\right|+3x=5\) `(1)`

Thường hợp `1` : `x-4 >= 0<=> x >=0` thì phương trình `(1)` thở thành :

`x-4 = 5-3x`

`<=> x+3x=5+4`

`<=> 4x=9`

`<=> x= 9/4 (t//m)`

Trường hợp `2` : `x-4< 0<=> x<0` thì phương trình `(1)` trở thành :

`-(x-4) =5-3x`

`<=> -x +4=5-3x`

`<=> -x+3x=5-4`

`<=> 2x =1`

`<=>x=1/2 ( kt//m)`

Vậy phương trình có nghiệm `x=9/4`

đây là phương trình mà đâu phải bất phương trình đâu

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

1:

a: =>3x=6

=>x=2

b: =>4x=16

=>x=4

c: =>4x-6=9-x

=>5x=15

=>x=3

d: =>7x-12=x+6

=>6x=18

=>x=3

2:

a: =>2x<=-8

=>x<=-4

b: =>x+5<0

=>x<-5

c: =>2x>8

=>x>4

2x(6x – 1) > (3x – 2)(4x + 3)

⇔ 12x2 – 2x > 12x2 – 8x + 9x – 6

⇔ 12x2 – 2x – 12x2 + 8x – 9x > -6 (Chuyển vế, đổi dấu)

⇔ -3x > -6

⇔ x < 2 (Chia cả hai vế cho -3 < 0, BPT đổi chiều)

Vậy bất phương trình có nghiệm x < 2.