a, \(x^2+4x=0\)
\(\Rightarrow x.\left(x+4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy......
b,\(x\left(3x-1\right)-5\left(1-3x\right)=0\)
\(\Rightarrow3x^2-x-5+15x=0\)
\(\Rightarrow3x^2+14x-5=0\)
\(\Rightarrow3x^2+15x-x-5=0\)
\(\Rightarrow\left(3x^2+15x\right)-\left(x+5\right)=0\)
\(\Rightarrow3x.\left(x+5\right)-\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right).\left(3x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+5=0\\3x-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy.....
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a ) \(x^2+4x=0\)
\(\Leftrightarrow x\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy ....................
b ) \(x\left(3x-1\right)-5\left(1-3x\right)=0\)
\(\Leftrightarrow x\left(3x-1\right)+5\left(3x-1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy...................
a) \(x^2+4x=0\\ < =>x\left(x+4\right)=0\\ =>\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
b) \(x\left(3x-1\right)-5\left(1-3x\right)=0\\ < =>x\left(3x-1\right)+5\left(3x-1\right)=0\\ < =>\left(x+5\right)\left(3x-1\right)=0\\ =>\left\{{}\begin{matrix}x+5=0\\3x-1=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)
a)x^2+4x=0
x*x+4x=0
x(x+4)=0
TH1:x=0 TH2:x+4=0
x= -4