Question

Va rog sa ma ajutati, d-au coroana, doresc sa fie scrisa pe o foaie de hartie (matematica) ​

Answer 1

Răspuns:

-1.29904

Explicație pas cu pas:

$\frac{4-\sqrt{24}}{\sqrt{32}}+\frac{6\sqrt{6}+9}{4\sqrt{27}}-\frac{6+4\sqrt{6}}{6\sqrt{12}}-\frac{8+\sqrt{24}}{3\sqrt{8}}=$

$=\frac{4-2\sqrt{6}}{4\sqrt{2}}+\frac{3\left(2\sqrt{6}+3\right)}{12\sqrt{3}}-\frac{2\left(3+2\sqrt{6}\right)}{12\sqrt{3}}-\frac{8+2\sqrt{6}}{6\sqrt{2}}=$

$=\frac{2\left(2-\sqrt{6}\right)}{4\sqrt{2}}+\frac{2\sqrt{6}+3}{4\sqrt{3}}-\frac{3+2\sqrt{6}}{6\sqrt{3}}-\frac{2\left(4+\sqrt{6}\right)}{6\sqrt{2}}=$

$=\frac{2-\sqrt{6}}{2\sqrt{2}}+\frac{\left(2\sqrt{6}+3\right)\sqrt{3}}{12}-\frac{\left(3+2\sqrt{6}\right)\sqrt{3}}{18}-\frac{4+\sqrt{6}}{3\sqrt{2}}=$

$=\frac{2\sqrt{2}-\sqrt{12}}{4}+\frac{6\sqrt{2}+3\sqrt{3}}{12}-\frac{3\sqrt{3}+6\sqrt{2}}{18}-\frac{4\sqrt{2}+\sqrt{12}}{6}=$

$=\frac{2\sqrt{2}-2\sqrt{3}}{4}+\frac{3\left(2\sqrt{2}+\sqrt{3}\right)}{12}-\frac{3\left(\sqrt{3}+2\sqrt{2}\right)}{18}-\frac{4\sqrt{2}+2\sqrt{3}}{6}=$

$=\frac{2\left(\sqrt{2}-\sqrt{3}\right)}{4}+\frac{2\sqrt{2}+\sqrt{3}}{4}-\frac{\sqrt{3}+2\sqrt{2}}{6}-\frac{2\left(2\sqrt{2}+\sqrt{3}\right)}{6}=$

$=\frac{\sqrt{2}-\sqrt{3}}{2}+\frac{2\sqrt{2}+\sqrt{3}}{12}-\frac{\sqrt{3}+2\sqrt{2}}{6}-\frac{2\sqrt{2}+\sqrt{3}}{3}=$

$=\frac{\sqrt{2}-\sqrt{3}}{2}-\frac{2\sqrt{2}+\sqrt{3}}{12}-\frac{\sqrt{3}+2\sqrt{2}}{6}=$

$=\frac{6\left(\sqrt{2}-\sqrt{3}\right)-\left(2\sqrt{2}+\sqrt{3}\right)-2\left(\sqrt{3}-2\sqrt{2}\right)}{12}=$

$=\frac{6\sqrt{2}-6\sqrt{3}-2\sqrt{2}-\sqrt{3}-2\sqrt{3}-4\sqrt{2}}{12}=$

$=\frac{0-9\sqrt{3}}{12}=$

$=\frac{-9\sqrt{3}}{12}=$

$=\frac{-3\sqrt{3}}{4}=-1.29904$

Answer 2

Răspuns:

Explicație pas cu pas: