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chị đã học lớp 8 rồi chắc chị sẽ biết toán lớp 5 hay giúp em nhe voi lai trang cua em la tat tien do
\(a,5x\left(x^2-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ b,3\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow3\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(3-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\\ c,x^2-9x-10=0\\ \Leftrightarrow x^2+x-10x-10=0\\ \Leftrightarrow x\left(x+1\right)-10\left(x+1\right)=0\\ \Leftrightarrow\left(x-10\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=10\end{matrix}\right.\)
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}
b: \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
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) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
(x - 2)(x2 + 2x + 4) + 2(x2 - 4) - 5(x - 2) = 0
(x - 2)(x + 2)2 + 2(x - 2)(x+2) - 5(x - 2) = 0
(x - 2)[(x+2)2 + 2(x+2) - 5]= 0
(x - 2)[(x + 2)2 + 2(x + 2) + 1 - 6] = 0
( x - 2)[(x + 2 + 1)2 - 6] = 0
(x - 2)[(x + 3)2 - 6] = 0
(x - 2)(x + 3 - \(\sqrt{6}\))(x + 3 + \(\sqrt{6}\)) = 0
TH1. x - 2 = 0 <=> x = 2
TH2. x + 3 - \(\sqrt{6}\) = 0 <=> x = \(\sqrt{6}-3\)
TH3. x + 3 + \(\sqrt{6}\) = 0 <=> x = \(-\sqrt{6}-3\)
S = {2; \(\sqrt{6}-3\); \(-\sqrt{6}-3\)}
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
\(x^2-x+0,25=0\)
\(\Leftrightarrow\)\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\)\(x-\frac{1}{2}=0\)
\(\Leftrightarrow\)\(x=\frac{1}{2}\)
Vậy x = 1/2
\(x^2-x+0.25=0\)
\(x^2-x+0=0\)
\(\Rightarrow x^2-x=0\)
\(\Rightarrow x=0\)