\(a,x^3-13x=0\)
\(x.\left(x^2-13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\sqrt{13}\end{cases}}}\)
\(b,2-25x^2=0\)
\(\Rightarrow25x^2=2\Rightarrow x^2=\frac{2}{25}\Rightarrow x=\sqrt{\frac{2}{25}}\)
\(c,x^2-x+\frac{1}{4}=0\)
\(\left(x-\frac{1}{2}\right)^2=0\Rightarrow x=\frac{1}{2}\)
a, x 3 - 13 x = 0
=> x ( x 2 - 13 ) = 0
=> \(\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow[\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)
b, 2 - 25 x 2 = 0
=> 25 x 2 = 2
=> x 2 = 0,08
=> \(\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)
x, x 2 - x + \(\frac{1}{4}\)= 0
=> \(\left(x-\frac{1}{2}\right)^2=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=\frac{1}{2}\)
ê, sorry nhá cái đoạn nì quên:
\(x=\pm\sqrt{13}\)
\(x=\pm\sqrt{\frac{2}{25}}\)
p/s: nguồn: Thăm Tuy Thăm Tuy sorry t quên >:
a) \(x^3-13x=0\Rightarrow x^3=13x\)
\(x^2=13\Rightarrow x=\sqrt{13}\)
b)\(2-25x^2=0\Rightarrow2=25x^2\)
\(\frac{2}{25}=x^2\Rightarrow x=\sqrt{\frac{2}{25}}\)
c) \(x^2-x+\frac{1}{4}=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
