/-3x+2/=0
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\(\left(3x-2\right)^2-6x+4=0\\ =>\left(3x-2\right)^2+2\left(-3x+2\right)=0\\ =>\left(2-3x\right)^2+2\left(2-3x\right)=0\\ =>\left(2-3x\right)\left(2-3x+2\right)=0\\ =>\left(2-3x\right)\left(4-3x\right)=0\\ \)
=> 2-3x=0 hoặc 4-3x=0
Nếu 2-3x=0 thì 3x=2 => \(x=\dfrac{2}{3}\)
Nếu 4-3x=0 thì 3x=4 => \(x=\dfrac{4}{3}\)
Vậy \(x=\dfrac{2}{3},x=\dfrac{4}{3}\)
a) \(\left(x-5\right)\left(2x-3^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\2x=9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{9}{2}\end{matrix}\right.\)
b) \(2\left(3x-15\right)\left(5-x\right)=0\)
\(\Rightarrow6\left(x-5\right)\left(5-x\right)=0\Rightarrow x=5\)
(x - 5)(2x - 32) = 0
=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)
\(a,\Rightarrow4x\left(x^2-9\right)=0\\ \Rightarrow4x\left(x-3\right)\left(x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\\ \Rightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Rightarrow2\left(x-3\right)4\left(x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
a) \(\Rightarrow4x\left(x^2-9\right)=0\)
\(\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow\left(3x-5-x-1\right)\left(3x-5+x+1\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(4x-4\right)=0\)
\(\Rightarrow8\left(x-3\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
\(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)
\(\Leftrightarrow3\left(x-2\right)\left(3-x\right)\left(3+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=2\end{matrix}\right.\)
a: \(x\left(1-2x\right)+2x^2=14\)
=>\(x-2x^2+2x^2=14\)
=>x=14
b: \(x\left(x-5\right)+3x-15=0\)
=>\(\left(x-5\right)\left(x+3\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left(x+1\right)^3=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
/-3x+2/=0
=> -3x+2=0
=> -3x=-2
=> x=-2:(-3)
=> x=2/3