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

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

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

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

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

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

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

\(\dfrac{x}{2x-6}-\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\left(ĐKXĐ:x\ne-1,x\ne3\right)\)

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

\(\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}-\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}\)

\(\Rightarrow x\left(x+1\right)-x\left(x-3\right)=4x\)

\(\Leftrightarrow x^2+x-x^2+3x=4x\)

\(\Leftrightarrow x^2+x-x^2+3x-4x=0\)

\(\Leftrightarrow0x=0\)

Phương trình có vô số nghiệm , trừ x = -1,x = 3

Vậy ...

\(\dfrac{12x+1}{12}< \dfrac{9x+1}{3}-\dfrac{8x+1}{4}\)

\(\Leftrightarrow12\cdot\dfrac{12x+1}{12}< 12\cdot\dfrac{9x+1}{3}-12\cdot\dfrac{8x+1}{4}\)

\(\Leftrightarrow12x+1< 4\left(9x+1\right)-3\left(8x+1\right)\)

\(\Leftrightarrow12x+1< 36x+4-24x-3\)

\(\Leftrightarrow12x+1< 12x+1\)

\(\Leftrightarrow12x-12x< 1-1\)

\(\Leftrightarrow0x< 0\)

Vậy S = {x | x \(\in R\)}

Ta có: | 2x | = x - 6

+ Với x ≥ 0, phương trình tương đương: 2x = x - 6 ⇔ x = - 6.

Không thỏa mãn điều kiện x ≥ 0.

+ Với x < 0, phương trình tương đương: - 2x = x - 6 ⇔ - 3x = - 6 ⇔ x = 2.

Không thỏa mãn điều kiện x < 0.

Vậy phương trình đã cho vô nghiệm.

Ta có: | 2x | = x - 6

+ Với x ≥ 0, phương trình tương đương: 2x = x - 6 ⇔ x = - 6.

Không thỏa mãn điều kiện x ≥ 0.

+ Với x < 0, phương trình tương đương: - 2x = x - 6 ⇔ - 3x = - 6 ⇔ x = 2.

Không thỏa mãn điều kiện x < 0.

Vậy phương trình đã cho vô nghiệm.

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

ĐKXĐ: \(x-2\ge0\Leftrightarrow x\ge2\)\(pt\Leftrightarrow\left[{}\begin{matrix}2x^2-7x+6=x-2\\2x^2-7x+6=2-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2-8x+8=0\\2x^2-6x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\left(x-2\right)^2=0\\2\left(x-2\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(N\right)\\x=1\left(L\right)\end{matrix}\right.\)

Vậy tập nghiệm của pt đó là \(S=\left\{2\right\}\)

`Đk:x>=2`

`+)2x^2-7x+6=x-2`

`<=>2x^2-4x-3x+6=x-2`

`<=>2x(x-2)-3(x-2)=x-2`

`<=>(x-2)(2x-3)=x-2`

`<=>(x-2)(2x-4)=0`

`<=>2(x-2)^2=0`

`=>x=2`

`+)2x^2-7x+6=2-x`

Tương tự như trên ta phân tích:`2x^2-7x+6=(x-2)(2x-3)`

`pt<=>(x-2)(2x-3)+(x-2)=0`

`<=>(x-2)(2x-2)=0`

Vì `x>=2=>2x-2>=2>0`

`=>x=2`

Vậy `S={2}`

`<

\(\Leftrightarrow2x^2+3x-4x-6-2\left(x^2-1\right)=6\)

\(\Leftrightarrow2x^2-x-6-2x^2+2-6=0\)

=>x+10=0

hay x=-10

\(\left(x-2\right)\left(2x+3\right)-2\left(x-1\right)\left(x+1\right)=6\\ \Rightarrow2x^2-4x+3x-6-2x^2+2-6=0\\ \Rightarrow-x-10=0\\ \Rightarrow-x=10\\ \Rightarrow x=-10\)

\(\left(x-2\right)\left(2x+3\right)-2\left(x-1\right)\left(x+1\right)=6\)

\(\Leftrightarrow2x^2-x-6-2\left(x^2-1\right)=6\Leftrightarrow-x=10\Leftrightarrow x=-10\)

5 – (6 – x) = 4(3 – 2x)

⇔ 5 – 6 + x = 12 – 8x

⇔ x + 8x = 12 – 5 + 6

⇔ 9x = 13

⇔ x = 13/9

Vậy phương trình có một nghiệm x = 13/9.