Question

Urgent!!!❤️ am nevoie de ajutor primesc nota

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\bf9$

$\it d)\ \sqrt{\dfrac{50}{3}}+ \sqrt{\dfrac{98}{3}} - \sqrt{\dfrac{242}{3}}= \sqrt{\dfrac{2}{3}}\Big(\sqrt{25}+\sqrt{49}-\sqrt{121}\Big)= \sqrt{\dfrac{2}{3}}\Big(5+7-11\Big)=\\ \\ \\ = \sqrt{\dfrac{2}{3}}\cdot1= \sqrt{\dfrac{2}{3}}=\dfrac{^{\sqrt3)}\sqrt2}{\ \sqrt3}= \dfrac{\sqrt6}{3}$

$\bf10.$

$\it c)\ \dfrac{\sqrt2}{4}-\dfrac{\sqrt{50}}{6}-\dfrac{\sqrt8}{3}+\dfrac{\sqrt{18}}{3}=\sqrt2\Big(\dfrac{1}{4}-\dfrac{5}{6}-\dfrac{2}{3}+\dfrac{3}{3}\Big)=\\ \\ \\ =\sqrt2\Big(\dfrac{^{3)}1}{\ 4}-\dfrac{^{2)}5}{\ 6}+\dfrac{^{4)}1}{\ 3}\Big)=\sqrt2\cdot \dfrac{3-10+4}{12}\sqrt2\cdot\dfrac{\ -3^{(3}}{12}=-\dfrac{\sqrt2}{4}$