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

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

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

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,\(2x+5=2-x\)

\(< =>2x+x+5-2=0\)

\(< =>3x+3=0\)

\(< =>x=-1\)

b, \(/x-7/=2x+3\)

Với \(x\ge7\)thì \(PT< =>x-7=2x+3\)

\(< =>2x-x+3+7=0\)

\(< =>x+10=0< =>x=-10\)( lọai )

Với \(x< 7\)thì \(PT< =>7-x=2x+3\)

\(< =>2x+x+3-7=0\)

\(< =>3x-4=0< =>x=\frac{4}{3}\) ( loại )

c,\(\frac{4}{x+2}-\frac{4x-6}{4x-x^3}=\frac{x-3}{x\left(x-2\right)}\left(đk:x\ne-2;0;2\right)\)

\(< =>\frac{4x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{4x-6}{x\left(x-2\right)\left(2+x\right)}=\frac{\left(x-3\right)\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)

\(< =>4x^2-8x+4x-6=x^2-x-6\)

\(< =>4x^2-x^2-4x+x-6+6=0\)

\(< =>3x^2-3x=0< =>3x\left(x-1\right)=0< =>\orbr{\begin{cases}x=0\left(loai\right)\\x=1\left(tm\right)\end{cases}}\)

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

1) \(2\left(x+3\right)>5\left(x-1\right)+2\Leftrightarrow2x+6>5x-5+2\Leftrightarrow3x>9\Leftrightarrow x>3\)

2) \(x^2-x\left(x+2\right)>3x-10\)

\(\Leftrightarrow x^2-x^2-2x>3x-10\Leftrightarrow5x< 10\Leftrightarrow x< 2\)

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

\(\Leftrightarrow x^2-5x< x^2+2x+1\Leftrightarrow7x>-1\Leftrightarrow x>-\dfrac{1}{7}\)

4) \(15-2\left(x-7\right)< 2\left(x-3\right)-6\)

\(\Leftrightarrow15-2x+14< 2x-6-6\Leftrightarrow4x>41\Leftrightarrow x>\dfrac{41}{4}\)

1: Ta có: \(2\left(x+3\right)>5\left(x-1\right)+2\)

\(\Leftrightarrow2x+6>5x-5+2\)

\(\Leftrightarrow-3x>-9\)

hay x<3

2: Ta có: \(x^2-x\left(x+2\right)>3x-10\)

\(\Leftrightarrow x^2-x^2-2x>3x-10\)

\(\Leftrightarrow-5x>-10\)

hay x<2

3: Ta có: \(x\left(x-5\right)\le\left(x+1\right)^2\)

\(\Leftrightarrow x^2-5x-x^2-2x-1\ge0\)

\(\Leftrightarrow-7x\ge1\)

hay \(x\le-\dfrac{1}{7}\)

4: Ta có: \(15-2\left(x-7\right)< 2\left(x-3\right)-6\)

\(\Leftrightarrow15-2x+14< 2x-6-6\)

\(\Leftrightarrow-4x< -12-29=-41\)

hay \(x>\dfrac{41}{4}\)

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Mình không biết sin lỗi vạn

a: =>x^2-8x+16<x^2-8x

=>16<0(loại)

b: =>\(x+\dfrac{1}{2}>=\dfrac{5x-3}{3}\)

=>x+1/2>=5/3x-1

=>-2/3x>=-3/2

=>x<=3/2:2/3=9/4

c: =>\(\dfrac{7-x}{4}< =\dfrac{2x-4}{3}\)

=>21-3x<=8x-16

=>-11x<=-37

=>x>=37/11

`|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`

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

Chọn A