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

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

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

Vậy \(S=\left\{3\right\}\)

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

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

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

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

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

\(\left(x-1\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(2x-3\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(3x-5\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\left(x-1\right)^2-\left(2x-3\right)\left(3x-5\right)\right)+\left(2x-3\right)\left(\left(2x-3\right)^2-\left(x-1\right)\left(3x-5\right)\right)+\left(3x-5\right)\left(\left(3x-5\right)^2-\left(x-1\right)\left(2x-3\right)\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(7-5x\right)+\left(2x-3\right)\left(x-2\right)^2+\left(3x-5\right)\left(x-2\right)\left(7x-11\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(\left(x-1\right)\left(7-5x\right)+\left(2x-3\right)\left(x-2\right)+\left(3x-5\right)\left(7x-11\right)\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(18x^2-63x+54\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\18x^2-63x+54=0\end{matrix}\right.\)

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

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

b. `|x + 1| + |2x - 3| = |3x - 2|`

Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)

Vậy phương trình có vô số nghiệm.

\(\Leftrightarrow\left(3x-5\right)^3-3\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(2x-3\right)^3+\left(x-1\right)^3=9\left(x-2\right)^2\left(2x-3\right)\)

\(\Rightarrow x^2-4x+4=0\)

\(\Rightarrow\left(-4\right)^2-4\left(1.4\right)=0\)(cái này là D )

\(\Rightarrow x_{1,2}=\frac{-b+-\sqrt{D}}{2a}=\frac{4+-\sqrt{0}}{2}\)

\(\Rightarrow2x-3=0\)

\(\Rightarrow2x=3\)

\(\Rightarrow x=\frac{3}{2}\)hoặc\(x=2\)

bạn cứ nhân từ từ ra rùi rút gọn là đc thui mà

3/2 k thỏa mãn nhé 

:v

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

a: =>9x^2+6x+1-6(2x^2-13x+21)=0

=>9x^2+6x+1-12x^2+78x-126=0

=>-3x^2+84x-125=0

=>\(x\in\left\{26.42;1.58\right\}\)

b: =>(3x+1)[(2x-5)^2-(x-3)^2]=0

=>(3x+1)(2x-5-x+3)(2x-5+x-3)=0

=>(3x+1)(x-2)(3x-8)=0

=>\(x\in\left\{-\dfrac{1}{3};2;\dfrac{8}{3}\right\}\)

c; =>(x+5)(0,75x-3+1,25x)=0

=>(x+5)(2x-3)=0

=>x=3/2 hoặc x=-5

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

\(\Leftrightarrow2x^2-10x-x^2-3x-3x-9=3x-5x+x^2\)

\(\Leftrightarrow2x^2-x^2-x^2-10x-3x-3x-3x+5x=-9\)

\(\Leftrightarrow-14x=-9\)

\(\Leftrightarrow x=-\dfrac{9}{14}\)

Vậy phương trình có nghiệm là \(S=\left\{-\dfrac{9}{14}\right\}\)

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

\(\Leftrightarrow2x^2-10x-\left(x^2+6x+9\right)-3x-5x+x^2=0\)

\(\Leftrightarrow2x^2-10x-x^2-6x-9-3x-5x+x^2=0\)

\(\Leftrightarrow2x^2-24x=0\)

\(\Leftrightarrow2x\left(x-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-12=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=12\end{matrix}\right.\)