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Tìm x, biết:
a, ( \(x\) - \(\dfrac{2}{3}\) )\(^2\) = \(\dfrac{5}{6}\)
b, ( \(\dfrac{3}{4}\) - \(x\) )\(^{^{ }3}\) = -8
Giúp mình vs nhaa, pls
a, \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{5}{6}=\left(\sqrt{\dfrac{5}{6}}\right)^2\)
TH1 : \(x-\dfrac{2}{3}=\sqrt{\dfrac{5}{6}}\Leftrightarrow x=\dfrac{2}{3}+\sqrt{\dfrac{5}{6}}\)
TH2 : \(x-\dfrac{2}{3}=-\sqrt{\dfrac{5}{6}}\Leftrightarrow x=-\sqrt{\dfrac{5}{6}}+\dfrac{2}{3}\)
b, \(\left(\dfrac{3}{4}-x\right)^3=-8=\left(-2\right)^3\Rightarrow\dfrac{3}{4}-x=-2\Leftrightarrow x=\dfrac{3}{4}+2=\dfrac{11}{4}\)
a, \(\left(x-\dfrac{2}{3}\right)^2=\dfrac{5}{6}=\left(\sqrt{\dfrac{5}{6}}\right)^2\)
TH1 : \(x-\dfrac{2}{3}=\sqrt{\dfrac{5}{6}}\Leftrightarrow x=\dfrac{2}{3}+\sqrt{\dfrac{5}{6}}\)
TH2 : \(x-\dfrac{2}{3}=-\sqrt{\dfrac{5}{6}}\Leftrightarrow x=-\sqrt{\dfrac{5}{6}}+\dfrac{2}{3}\)
b, \(\left(\dfrac{3}{4}-x\right)^3=-8=\left(-2\right)^3\Rightarrow\dfrac{3}{4}-x=-2\Leftrightarrow x=\dfrac{3}{4}+2=\dfrac{11}{4}\)