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

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

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

Vậy \(S=\left\{3\right\}\)

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

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

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

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

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(a,PT\Leftrightarrow8x^3-6x^2+4x-3=3x^3-36x^2+x-12\)

\(\Leftrightarrow5x^3+30x^2+3x+9=0\)

\(\Leftrightarrow x=-5,95...\)

\(b,PT\Leftrightarrow2x+22-3x^2-33x=6x-15x^2-4+10x\)

\(\Leftrightarrow12x^2-47x+26=0\)

<=> (3x - 2)(4x - 13) = 0

<=> x = 2/3 hoặc x = 13/4

c, Tách ra <=> (2x - 1)(2x - 5) = 0 <=> ...

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

<=> 4x² - 1 - ( x² - 2x + 1 ) - 3x( x - 2 ) = 0

<=> 4x² - 1 - x² + 2x - 1 - 3x² + 6x = 0

<=> 8x - 2 = 0

<=> x = 1/4

Vậy phương trình có 1 nghiệm x = 1/4

b) ( 4x - 3 )( 3x + 2 ) = 2( 3x - 1 )( 2x + 5 )

<=> 12x² - x - 6 - 2( 6x² + 13x - 5 ) = 0

<=> 12x² - x - 6 - 12x² - 26x + 10 = 0

<=> -27x + 4 = 0

<=> x = 4/27

Vậy phương trình có 1 nghiệm x = 4/27

c) ( x - 1 )( x² + x + 1 ) - 5( 2x - 3 ) = x( x² - 3 )

<=> x³ - 1 - 10x + 15 - x( x² - 3 ) = 0

<=> x³ + 14 - 10x - x³ + 3x = 0

<=> -7x + 14 = 0

<=> x = 2

Vậy phương trình có nghiệm x = 2

d) \(\frac{3x-2}{4}-\frac{x+4}{3}=\frac{1+x}{12}\)

<=> \(\frac{3x}{4}-\frac{2}{4}-\frac{x}{3}-\frac{4}{3}=\frac{1}{12}+\frac{x}{12}\)

<=> \(\frac{3}{4}x-\frac{1}{3}x-\frac{1}{12}x=\frac{1}{12}+\frac{1}{2}+\frac{4}{3}\)

<=> \(x\left(\frac{3}{4}-\frac{1}{3}-\frac{1}{12}\right)=\frac{23}{12}\)

<=> \(x\cdot\frac{1}{3}=\frac{23}{12}\)

<=> x = 23/4

Vậy phương trình có 1 nghiệm x = 23/4

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,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)

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

c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)

TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)

TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)

Vậy x=0,5...

d, đk \(x\ge-1\)

=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)

\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)

a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow\left|3x-1\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

b) Ta có: \(\sqrt{4x^2-4x+1}=3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)

\(\Leftrightarrow\left|x-3\right|=4-3x\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)

1. 4x-12=0

<=>4x=12

<=>x=3

2.  x.(x+1)-(x+2)(x+3)=7

<=>x²+x-x²-3x-2x-6=7

<=>x²-x²+x-2x-3x=7+6

<=>-4x=13

<=>x=\(-\dfrac{13}{4}\)

3.   7+2x=22-3x

<=>2x+3x=22-7

<=>5x=15

<=>x=3

4.  (x-1)-(2x-1)=9-x

<=>x-1-2x+1=9-x

<=>x-2x+x=9+1-1

<=>0x=9

vô nghiệm

1.

$\sqrt{3x^2}-\sqrt{12}=0$

$\Leftrightarrow \sqrt{3x^2}=\sqrt{12}$

$\Leftrightarrow 3x^2=12$

$\Leftrightarrow x^2=4$

$\Leftrightarrow (x-2)(x+2)=0\Leftrightarrow x=\pm 2$

2. 

$\sqrt{(x-3)^2}=9$

$\Leftrightarrow |x-3|=9$

$\Leftrightarrow x-3=9$ hoặc $x-3=-9$

$\Leftrightarrow x=12$ hoặc $x=-6$

3.

$\sqrt{4x^2+4x+1}=6$

$\Leftrightarrow \sqrt{(2x+1)^2}=6$

$\Leftrightarrow |2x+1|=6$

$\Leftrightarrow 2x+1=6$ hoặc $2x+1=-6$

$\Leftrightarrow x=\frac{5}{2}$ hoặc $x=\frac{-7}{2}$

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