Question

Repedeee răspuns dau maxima mea de ou cte plus celai istet​

Answer 1

Explicație pas cu pas:

a)

$\frac{^{ \sqrt{3} )}12}{ \sqrt{3} } + \frac{^{ \sqrt{3} )}15}{5 \sqrt{3} } - \frac{^{ \sqrt{3} )}21}{7 \sqrt{3} } + \frac{^{ \sqrt{3} )}18}{ \sqrt{3} } = \frac{12 \sqrt{3} }{3} + \frac{15 \sqrt{3} }{15} - \frac{21 \sqrt{3} }{21} + \frac{18 \sqrt{3} }{3} = 4 \sqrt{3} + \sqrt{3} - \sqrt{3} + 6 \sqrt{3} = \bf 10 \sqrt{3}$

b)

$\frac{^{4)}8}{3 \sqrt{2} } - \frac{^{6)}8}{2 \sqrt{2} } + \frac{^{2)}8}{6 \sqrt{2} } - \frac{8}{12 \sqrt{2} } = \frac{32}{12 \sqrt{2}} - \frac{48}{12 \sqrt{2}} + \frac{16}{12 \sqrt{2}} - \frac{8}{12 \sqrt{2}} = \frac{32 - 48 + 16 - 8}{12 \sqrt{2} } = - \frac{8}{12 \sqrt{2} } = - \frac{8 \sqrt{2} }{24} = \bf - \frac{ \sqrt{2} }{3}$

c)

$\frac{^{2)} 24}{5 \sqrt{6} } - \frac{^{10)}12}{ \sqrt{6} } + \frac{21}{10 \sqrt{6} } - \frac{^{5)}28}{2 \sqrt{6} } = \frac{48}{10 \sqrt{6}} - \frac{120}{10\sqrt{6} } + \frac{21}{10 \sqrt{6} } - \frac{140}{10 \sqrt{6} } = \frac{48 - 120 + 21 - 140}{10 \sqrt{6} } = - \frac{191}{10 \sqrt{6} } = \bf - \frac{191 \sqrt{6} }{60}$

d)

$\frac{^{15)}7}{2 \sqrt{5} } + \frac{^{10)}7}{3 \sqrt{5} } - \frac{^{5)}7}{6 \sqrt{5} } + \frac{^{6)}7}{5 \sqrt{5} } = \frac{7(15 + 10 - 5 + 6)}{30 \sqrt{5} } = \frac{7 \cdot 26}{30 \sqrt{5} } = \bf \frac{91 \sqrt{5} }{75}$