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1) (x-1)(x+5)(-3x+8)=0
\(\hept{\begin{cases}\\\\\end{cases}}\)
1) (x-1)(x+5)(-3+8)=0
= (x-1)(x+5).5 =0
\(\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0+1=1\\x=0-5=-5\end{cases}}\)
\(\Rightarrow x\in\left\{1;-5\right\}\)
2) (x-1)(x-2)(x-3)=0
\(\hept{\begin{cases}x-1=0\\x-2=0\\x-3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0+1=1\\x=0+2=2\\x=0+3=3\end{cases}}\)
\(\Rightarrow x\in\left\{1;2;3\right\}\)
3)(5x+3)(x2+4)(x-1)=0
\(\hept{\begin{cases}5x+3=0\\x^2+4=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}5x=0-3=-3\\x^2=0-4=-4\\x=0+1=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-3:5\Rightarrow x\in\varnothing\\x\in\varnothing\\x=1\end{cases}}\)
\(\Rightarrow x=1\)
4)x(x2-1)=0
\(\orbr{\begin{cases}x=0\\x^2-1=0\Rightarrow x^2=0+1=1\Rightarrow x^2=1^2;(-1)^2\Rightarrow x\in\left\{1;-1\right\}\end{cases}}\)
\(\Rightarrow x\in\left\{-1;0;1\right\}\)
Xin lỗi về phần bên trên nha! tại tui ấn nhầm nút.Sorry.

a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)
\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)
b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)
\(3x-\frac{3}{2}-5x-3=x+\frac{1}{5}\)
\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)
\(-3x=\frac{47}{10}\)
\(x=\frac{-47}{30}\)
c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)
\(-7x=\frac{-1}{6}\)
\(x=\frac{1}{42}\)
d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)
\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)
\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(3x=\frac{1}{6}\)
\(x=\frac{1}{18}\)
Học tốt nhé bn!

Ta có : \(\left(3-x\right)\left|x+5\right|=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)
Ta có :
\(\left(3-x\right)\left|x+5\right|=0\)
\(\Rightarrow\orbr{\begin{cases}3-x=0\\\left|x+5\right|=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

\(a,\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(\frac{11}{15}x=\frac{2}{5}\)
\(x=\frac{6}{11}\)
b,\(\left(2x-3\right).\left(6-2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\6-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=3\end{cases}}\)
Vậy

1/ ( x+12)(3-x)=0
=> \(\orbr{\begin{cases}x+12=0\\3-x=0\end{cases}}\)
=>\(\orbr{\begin{cases}x=-12\\x=3\end{cases}}\)

Làm theo công thức: tích bằng 0 thì một trong x thừa số bằng 0 rồi xét các trường hợp

\(1,x.\left(x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)
\(2,\left(x+12\right).\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)
\(3,\left(-x+5\right).\left(3-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)
4/ \(x.\left(2+x\right).\left(7-x\right)=0\)
\(\hept{\begin{cases}x=0\\2+x=0\\7-x=0\end{cases}}\) => \(\hept{\begin{cases}x=0\\x=-2\\x=7\end{cases}}\)
Vậy \(x=\left\{0,-2,7\right\}\)
5/ \(\left(x-1\right).\left(x+2\right).\left(-x-3\right)=0\)
\(\hept{\begin{cases}x-1=0\\x+2=0\\-x-3=0\end{cases}}\)=> \(\hept{\begin{cases}x=1\\x=-2\\x=-3\end{cases}}\)

Ta có: 4 - |x - 5| = 0
=> | x - 5 | = 4
<=> x - 5 = 4
x - 5 = -4
<=> x = 4 + 5
x = -4 + 5
<=> x = 9
x = 1
(x - 1/3).(1/2 - 3x) = 0
TH1:
x - 1/3 = 0
x = 1/3
TH2
1/2 - 3x = 0
3x = 0 - 1/2
3x = -1/2
x = -1/2 : 3
x = -1/6
Vậy x ϵ { - 1/6 ; 1/2}