a: \(\left(x-5\right)\left(x+3\right)\left(3x-3\right)=0\)
=>\(3\left(x-5\right)\left(x+3\right)\left(x-1\right)=0\)
=>(x-5)(x+3)(x-1)=0
=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\\x=1\end{matrix}\right.\)
b: \(9x^2-25=0\)
=>\(9x^2=25\)
=>\(x^2=\dfrac{25}{9}\)
=>\(\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
c: \(36x^2-6x\left(x-1\right)=0\)
=>\(6x\left(6x-x+1\right)=0\)
=>6x(5x+1)=0
=>x(5x+1)=0
=>\(\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
d: \(\left(x-5\right)^2=\left(2x+1\right)^2\)
=>(2x+1-x+5)(2x+1+x-5)=0
=>(x+6)*(3x-4)=0
=>\(\left[{}\begin{matrix}x+6=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{3}\end{matrix}\right.\)
$#flo$
`a) (x-5).(x+3).(3x-3) = 0`
`=> 3(x-5)(x+3)(x-3) = 0`
`=> (x-5)(x+3)(x-3) = 0`
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x+3=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\\x=3\end{matrix}\right.\)
Vậy `x`∈`{5;-3;3}`
`b)9x^2 - 25 = 0`
`=> (3x)^2 - 5^2 = 0`
`=> (3x-5)(3x + 5) = 0`
\(\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=5\\3x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
Vậy `x`∈`{5/3 ; -5/3}`
`c)36x^2 - 6x. (x-1) = 0`
`=> 6x(6x - x + 1) = 0`
\(\Rightarrow\left[{}\begin{matrix}6x=0\\6x-x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\5x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy `x`∈`{0;-1/5}`
`d)(x-5)^2 = (2x+1)^2`
`=> (x-5)^2 - (2x + 1)^2 = 0`
`=> (x-5-2x-1)(x-5 + 2x + 1) = 0`
\(\Rightarrow\left[{}\begin{matrix}-x-6=0\\3x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{3}\end{matrix}\right.\)
Vậy `x`∈`{-6 ; 4/3}`
