\(\dfrac{x+8}{12}+\dfrac{x+9}{11}+\dfrac{x+10}{10}+3=0\\ \Leftrightarrow\dfrac{x+8}{12}+1+\dfrac{x+9}{11}+1+\dfrac{x+10}{10}+1=0\\ \Leftrightarrow\dfrac{x+20}{12}+\dfrac{x+20}{11}+\dfrac{x+20}{10}=0\\ \Leftrightarrow\left(x+20\right)\left(\dfrac{1}{12}+\dfrac{1}{11}+\dfrac{1}{10}\right)=0\\ \Leftrightarrow x+20=0\Leftrightarrow x=-20\\ KL:...\)

`<=>((x+8)/12+1)+((x+9)/11+1)+((x+10)/10+1)=0`

`<=>(x+20)/12+(x+20)/11+(x+20)/10=0`

`<=>(x+20)(1/12+1/11+1/10)=0`

Vì `1/12+1/11+1/10 ≠ 0`

`=>x+20=0`

`=>x=0-20`

`=>x=-20`

Ta có :\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

=> \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

=> \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

=> \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

=> x + 20 = 0

=> x = -20

Vậy x = -20

\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

\(\Leftrightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

\(\Rightarrow x+20=0\Rightarrow x=-20\)

cái bên vế phải chả hiểu là j ?? Tớ tạm đặt là    \(\alpha\)    nhé      (   \(\alpha\)là 1 hằng số    )

pt <=>   \(\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=\alpha\)

<=>   \(\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=\alpha\)

<=>   \(\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=\alpha\)

<=>   \(\frac{\left(x+20\right).181}{660}=\alpha\)

<=>   \(x=\frac{660\alpha}{181}-20\)

Bạn tự thay    \(\alpha\)là 1 số nào tùy ý cũng sẽ tìm ngay ra x thôi 

\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

\(\Leftrightarrow\frac{x+8}{12}+1+\frac{x+9}{11}+1+\frac{x+10}{10}+1=0\)

\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\right)=0\Leftrightarrow x=-20\)

x+812812+x+911911+x+10101010+3=0

⇒x+2323+x+ 911911+x+1+3=0

(x+x+x)+( 2323+ 911911+1+3)=0  

                                        3x+1813318133=0

                                       3x =0-1813318133 

                                       3x =−18133−18133 

                                       x =−18133−18133:3

                                      x =−18199−18199 

7)(16-8x)(2-6x)=0  

=> 16 - 8x = 0 hoặc 2 - 6x = 0

=> 16 = 8x hoặc 2 = 6x

=> x = 2 hoặc x = 1/3
8) (x+4)(6x-12)=0  

=> x + 4 = 0 hoặc 6x - 12 = 0

=> x = -4 hoặc x = 2
9) (11-33x)(x+11)=0 

=> 11 - 33x = 0 hoặc x + 11 = 0

=> x = 1/3 hoặc x = -11
10) (x-1/4)(x+5/6)=0 

=> x - 1/4 = 0 hoặc x + 5/6 = 0

=> x = 1/4 hoặc x = -5/6
11) (7/8-2x)(3x+1/3)=0  

=> 7/8 - 2x = 0 hoặc 3x + 1/3 = 0

=> 2x = 7/8 hoặc 3x = -1/3

=> x = 7/16 hoặc x = -1/9
12)3x-2x^2=0  

=> x(3 - 2x) = 0

=> x = 0 hoặc 3 - 2x = 0

=> x = 0 hoặc x = 3/2

\(a,\left(16-8x\right)\left(2-6x\right)=0\)

\(\hept{\begin{cases}16-8x=0\\2-6x=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=\frac{1}{3}\end{cases}}}\)

\(b,\left(x+4\right)\left(6x-12\right)=0\)

\(\hept{\begin{cases}x+4=0\\6x-12=0\end{cases}\Rightarrow\hept{\begin{cases}x=-4\\x=2\end{cases}}}\)

\(c,\left(11-33x\right)\left(x+11\right)=0\)

\(\hept{\begin{cases}11-33x=0\\x+11=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\\x=-11\end{cases}}}\)

\(d,\left(x-\frac{1}{4}\right)\left(x+\frac{5}{6}\right)=0\)

\(\hept{\begin{cases}x-\frac{1}{4}=0\\x+\frac{5}{6}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{5}{6}\end{cases}}}\)

\(e,\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\hept{\begin{cases}\frac{7}{x}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{7}{4}\\x=-\frac{1}{9}\end{cases}}}\)

\(f,3x-2x^2=0\)

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

\(\hept{\begin{cases}x=0\\3-2x=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}}\)

mk lưu nhầm ảnh ở bài dưới của câu

\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

\(\Rightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)\

\(\Rightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

\(\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)\ne0\Rightarrow x+20=0\Leftrightarrow x=0-20=-20\)

\(\frac{x+8}{12}+\frac{x+9}{11}+\frac{x+10}{10}+3=0\)

\(\Leftrightarrow\left(\frac{x+8}{12}+1\right)+\left(\frac{x+9}{11}+1\right)+\left(\frac{x+10}{10}+1\right)=0\)

\(\Leftrightarrow\frac{x+8+12}{12}+\frac{x+9+11}{11}+\frac{x+10+10}{10}=0\)

\(\Leftrightarrow\frac{x+20}{12}+\frac{x+20}{11}+\frac{x+20}{10}=0\)

\(\Leftrightarrow\left(x+20\right)\left(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\right)=0\)

Vì \(\frac{1}{12}+\frac{1}{11}+\frac{1}{10}\ne0\)

=> \(x+20=0\)

=> \(x=-20\)