tìm x, biết 3x - x2 = 0
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a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
a, 5\(x\)(\(x^2\) - 9) = 0
\(\left[{}\begin{matrix}x=0\\x^2-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy \(x\) \(\in\) { -3; 0; 3}
b, 3.(\(x+3\)) - \(x^2\) - 3\(x\) = 0
3.(\(x+3\)) - \(x\).( \(x\) + 3) = 0
(\(x+3\))( 3 - \(x\)) = 0
\(\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -3; 3}
c, \(x^2\) - 9\(x\) - 10 = 0
\(x^2\) + \(x\) - 10\(x\) - 10 = 0
\(x.\left(x+1\right)\) - 10.( \(x-1\)) = 0
(\(x+1\))(\(x-10\)) = 0
\(\left[{}\begin{matrix}x+1=0\\x-10=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
Vậy \(x\) \(\in\){ -1; 10}
\(\Rightarrow2\left(x+3\right)-x\left(x+3\right)=0\Rightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
b) 2(x + 3) – x2 – 3x = 0
⇔ 2(x + 3) - x(x + 3) = 0
⇔ (x + 3)(2 - x) = 0
⇔ x = -3 hoặc x = 2
Vậy: x = -3; x = 2
có 2 TH
TH1 3x=0 suy ra x=0:3 suy ra x=0
TH2 x2=0 suy ra x2=02 suy ra x=0
gia khánh lm sai ~~
Dấu (-) thì lm s chia trường hợp được
\(3x-x^2=0\)
\(\Rightarrow x\left(3-x\right)=-x\left(x-3\right)\)
\(Th1:-x=0\)
\(th2:x-3=0\Rightarrow x=3\)