d) \(x^2=a\left(a\ge0\right)\)
\(\Rightarrow x=\sqrt{a}\)
e) \(x^2=\dfrac{4}{9}\)
\(\Rightarrow x^2=\left(\pm\dfrac{2}{3}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
f) \(x^2-\dfrac{16}{25}=0\)
\(\Rightarrow x^2=\dfrac{16}{25}\)
\(\Rightarrow x^2=\left(\pm\dfrac{4}{5}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
g) \(x^2-\dfrac{7}{36}=0\)
\(\Rightarrow x^2=\dfrac{7}{36}\)
\(\Rightarrow x^2=\left(\pm\sqrt{\dfrac{7}{36}}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{7}{36}}\\x=-\sqrt{\dfrac{7}{36}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{7}}{6}\\x=-\dfrac{\sqrt{7}}{6}\end{matrix}\right.\)
h) Ta có: \(x^2\ge0\forall x\)
\(\Rightarrow x^2+1\ge1>0\forall x\)
mà \(x^2+1=0\)
nên không tìm được giá trị nào của x thoả mãn đề bài.