\(a,\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-1\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)
\(c,\Leftrightarrow\left(x-1\right)^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-4\\x-1=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\\ d,\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\\ \Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
a)25x2-4=0
⇔(5x)2-22=0
⇔(5x-2)(5x+2)=0
⇔5x-2=0 hay 5x+2=0
⇔5x=2 hay 5x=-2
⇔x=\(\dfrac{2}{5}\) hay x=\(\dfrac{-2}{5}\)
b)2x(x+5)-x-5=0
⇔2x(x+5)-(x+5)=0
⇔(x+5)(2x-1)=0
⇔x+5=0 hay 2x-1=0
⇔x=-5 hay 2x=1
⇔x=-5 hay x=\(\dfrac{1}{2}\)
c)x2-2x+1=0
⇔(x-1)2=0
⇔x-1=0
⇔x=1
d)x2-x-6=0
⇔x2-3x+2x-6=0
⇔(x2-3x)+(2x-6)=0
⇔x(x-3)+2(x-3)=0
⇔(x-3)(x+2)=0
⇔x-3=0 hay x+2=0
⇔x=3 hay x=-2
giúp em với mn ơi,e đag cần gấp, mn giúp em với ạ
giúp em với ạ, em đang vội mong mn giúp đỡ
