a)5x(x-1)=x-1
<=>5x(x-1)-(x-1)=0
<=>(x-1)(5x-1)=0
<=>x=1 hoặc x=1/5
a) \(5x\left(x-1\right)=x-1\)
\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=\dfrac{1}{5}\)
b) \(2\left(x+5\right)-x^2-5x=0\)
\(\Leftrightarrow2\left(x+5\right)-\left(x^2+5x\right)=0\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(x=-5\) hoặc \(x=2\)
c) \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left[x^2-\left(\dfrac{1}{2}\right)^2\right]=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)
b)2(x + 5) - x² - 5x = 0
<=> 2x + 10 - x² - 5x = 0
<=> -x² - 3x + 10 = 0
<=> x = -5
<=> x = 2w2
C) x³ -x/4 = 0 <=> x(x² - 1/4) = 0 <=> x(x-1/2)(x+1/2) = 0
VT là tích của 3 thừa số, VT là 0 nên ta có:
x = 0 hoặc (x-1/2) = 0 hoặc (x+1/2) = 0
<=> x = 0 hoặc x = 1/2 hoặc x = -1/2
a) 5x(x-1) = x-1 <=> 5x(x-1) - (x-1) =0 <=> (x-1)(5x-1) =0
<=> \(\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) 2(x+5) - x^2 -5x =0 <=> 2(x+5) - x(x+5) =0 <=> (x+5)(2-x) =0
<=>\(\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
c) x^3 -1/4x =0 <=> x(x^2 -1/4) =0 <=> x(x-1/2)(x+1/2) =0
<=> \(\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
