Question

Calculati:
Ofer coroana!​

Answer 1

Am atasat o rezolvare.

Answer 2

$\it \dfrac{6}{\sqrt2}[\sqrt{0,(8)}-\sqrt{12,5}]+(-\sqrt2)^6:[\sqrt{0,(6)}]^2= ?\\ \\ \\\sqrt{0,(8)}=\sqrt{\dfrac{8}{9}}=\sqrt{\dfrac{4\cdot2}{9}}=\dfrac{2\sqrt2}{3}\\ \\ \\ \sqrt{12,5}=\sqrt{\dfrac{25}{2}}=\dfrac{5}{\sqrt2}\\ \\ \\ (-\sqrt2)^6=[(-\sqrt2)^2]^3=2^3=8$

$\it [\sqrt{0,(6)}]^2=0,(6) =\dfrac{6^{(3}}{9}=\dfrac{2}{3}$

Acum exercițiul se poate scrie:

$\it \dfrac{6}{\sqrt2}\Big(\dfrac{2\sqrt2}{3}-\dfrac{5}{\sqrt2}\Big)+8:\dfrac{2}{3}=\dfrac{6}{\sqrt2}\cdot\dfrac{2\sqrt2}{3}-\dfrac{6}{\sqrt2}\cdot\dfrac{5}{\sqrt2} +8\cdot\dfrac{3}{2}=\\ \\ \\ =4-15+12=16-15=1$