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) Ta có: \(3x-1=0\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

Vậy: \(S=\left\{\dfrac{1}{3}\right\}\)

b) Ta có: \(5x-2=x+4\)

\(\Leftrightarrow5x-x=4+2\)

\(\Leftrightarrow4x=6\)

\(\Leftrightarrow x=\dfrac{3}{2}\)

Vậy: \(S=\left\{\dfrac{3}{2}\right\}\)

\(\text{ 2x+6=0 }\)

\(\Leftrightarrow2x=-6\)

\(\Leftrightarrow x=-3\)

\(S=\left\{-3\right\}\)

\(\text{3x-9=0 }\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\)

\(S=\left\{3\right\}\)

\(\text{4x+20=0}\)

\(\Leftrightarrow4x=-20\)

\(\Leftrightarrow x=-5\)

\(S=\left\{-5\right\}\)

\(\text{4x+1=6-x}\)

\(\Leftrightarrow4x+1-6-x=0\)

\(\Leftrightarrow3x-5=0\)

\(\Leftrightarrow3x=5\)

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

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

a: 2x+6=0

=>2x=-6

=>x=-3

b: 3x-9=0

=>3x=9

=>x=3

c: 4x+20=0

=>x+5=0

=>x=-5

d: 4x+1=6-x

=>5x=5

=>x=1

\(4x+1=6-x\)

\(\Leftrightarrow4x+1-6+x=0\)

\(\Leftrightarrow5x-5=0\)

\(\Leftrightarrow5x=5\)

\(\Leftrightarrow x=1\)

\(S=\left\{1\right\}\)

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

\(\frac{5\left(5-x\right)}{15}+\frac{3\left(3x-2\right)}{15}=\frac{2\left(2x+2\right)-1}{2\left(3-x\right)}\)

\(25-x+9x-6=\frac{2x+1}{3-x}\)

\(19+8x=\frac{2x+1}{3-x}\)

\(19+8x=\left(2x+1\right).\frac{1}{3-x}\)

\(19+8x=\frac{2x}{3-x}+\frac{1}{3-x}\)

\(19+8x=2x+1\)

\(19+8x-2x-1=0\)

\(18+6x=0\Leftrightarrow6x=-18\Leftrightarrow x=-3\)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

a: Ta có: \(3x+5\le4x-9\)

\(\Leftrightarrow-x\le-14\)

\(\Leftrightarrow x\ge14\)

b: Ta có: \(6-2x< 6-x\)

\(\Leftrightarrow-x< 0\)

hay x>0

c: Ta có: \(7\left(x-1\right)+5>-3x\)

\(\Leftrightarrow7x-7+5+3x>0\)

\(\Leftrightarrow10x>2\)

hay \(x>\dfrac{1}{5}\)

đặt \(\sqrt{3x^2+x+2}=a\)

\(a^2+4x^2+x^2-4x+4\)=4ax <=> \(\left(a^2-4ax+4x^2\right)+\left(x^2-4x+4\right)\)=0 <=>(a-2x)²+(x-2)²=0 

=>a=2x và x=2 đồng thởi xảy ra (1)

với x=2 =>a=\(\sqrt{3.4+2+2}\)=4=2x

vậy x=2 thỏa mãn điều kiện (1) =>pt co nghiệm duy nhất x=2

\(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}-\dfrac{2}{x^2-4x+3}\) = 0  (ĐKXĐ: x \(\ne\) 1; x \(\ne\) 2; x \(\ne\) 3)

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

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

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

Vậy pt trên có vô số nghiệm và x \(\ne\) 1; x \(\ne\) 2; x \(\ne\) 3

Chúc bn học tốt!

3x+5

=0

\(pt\text{ tương đương:}\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}=0\text{ tương đương với: }\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}=0\text{ hay:}\dfrac{1}{x-1}-\dfrac{1}{x-3}-\dfrac{1}{x-1}+\dfrac{1}{x-3}=0\text{ đúng}\)