3 years ago

Can somebody help me?

Mathematics

2 answers:

Cloud [144]3 years ago

6 0

Answer:

3rd one

Step-by-step explanation:

Marta_Voda [28]3 years ago

6 0

The answer is the third one

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Neporo4naja [7]

$x^2=\dfrac{2}{3}\to x=\pm\sqrt{\dfrac{2}{3}}\\\\x=-\dfrac{\sqrt{2}}{\sqrt3}\ \vee\ x=\dfrac{\sqrt2}{\sqrt3}\\\\x=-\dfrac{\sqrt2\cdot\sqrt3}{\sqrt3\cdot\sqrt3}\ \vee\ x=\dfrac{\sqrt2\cdot\sqrt3}{\sqrt3\cdot\sqrt3}\\\\\boxed{x=-\dfrac{\sqrt6}{3}\ \vee\ x=\dfrac{\sqrt6}{3}}$