\(a,x^2=9\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(b,x^2=10\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}\\x=-\sqrt{10}\end{matrix}\right.\)
\(c,2x^2+5=0\Leftrightarrow x^2=-\dfrac{5}{2}\) (vô lí vì \(x^2\ge0\))
Vậy \(x\in\varnothing\)
\(d,4-2x^2=0\\ \Leftrightarrow x^2=\dfrac{-4}{-2}=2\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
a: Ta có: \(x^2=9\)
nên \(x\in\left\{3;-3\right\}\)
b: Ta có: \(x^2=10\)
nên \(x\in\left\{\sqrt{10};-\sqrt{10}\right\}\)
c: Ta có: \(2x^2+5=0\)
\(\Leftrightarrow2x^2=-5\)(vô lý)
d: Ta có: \(4-2x^2=0\)
\(\Leftrightarrow2x^2=4\)
\(\Leftrightarrow x^2=2\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
