Question

alguien me ayuda es de matematica suma algebraicas

$\sqrt{5} + \sqrt{8} - \sqrt{32}$
$3 \times \sqrt{7} - 3 \times \sqrt{28} + \sqrt{63}$
$- 4 \times \sqrt{ \frac{1}{27} } + \sqrt{ \frac{1}{3} } - 2 \times \sqrt{ \frac{1}{243} }$
$\sqrt{54} + \sqrt{12} - \sqrt{6}$
$\sqrt{20} + 3 \times \sqrt{8} - 5 \times \sqrt{5}$
$- 3 \times \sqrt{ \frac{1}{2} } - 5 \times \sqrt{ \frac{1}{32} } + \sqrt{ \frac{1}{8} }$

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Answer 1

Respuesta:

Explicación paso a paso:

$\sqrt{5}+\sqrt{8}-\sqrt{32} \\\\=\sqrt{5}+2\sqrt{2}-4\sqrt{2}\\\\=\sqrt{5}-2\sqrt{2}$

$3\sqrt{7}-3\sqrt{28}+\sqrt{63}\\\\=3\sqrt{7}-6\sqrt{7}+3\sqrt{7}\\\\=0$

$-4\sqrt{\frac{1}{27}}+\sqrt{\frac{1}{3}}-2\sqrt{\frac{1}{243}}\\\\=-\frac{4}{3\sqrt{3}}+\sqrt{\frac{1}{3}}-\frac{2}{9\sqrt{3}}\\\\=\frac{-12-2}{9\sqrt{3}}+\sqrt{\frac{1}{3}}\\\\=\frac{-14}{9\sqrt{3}}+\sqrt{\frac{1}{3}}\\\\=\sqrt{\frac{1}{3}}-\frac{14}{9\sqrt{3}}\\\\=\sqrt{\frac{1}{3}}-\frac{14\sqrt{3}}{27}$

$\sqrt{54}+\sqrt{12}-\sqrt{6}\\\\=3\sqrt{6}+2\sqrt{3}-\sqrt{6}\\\\=2\sqrt{6}+2\sqrt{3}$

$\sqrt{20}+3\sqrt{8}-5\sqrt{5}\\\\=2\sqrt{5}+6\sqrt{2}-5\sqrt{5}\\\\=-3\sqrt{5}+6\sqrt{2}$

$-3\sqrt{\frac{1}{2}}-5\sqrt{\frac{1}{32}}+\sqrt{\frac{1}{8}}\\\\=-3\sqrt{\frac{1}{2}}-\frac{5}{4\sqrt{2}}+\frac{1}{2\sqrt{2}}\\\\=-\frac{5}{4\sqrt{2}}+\frac{2}{4\sqrt{2}}-3\sqrt{\frac{1}{2}}\\\\=\frac{-5+2}{4\sqrt{2}}-3\sqrt{\frac{1}{2}}\\=\frac{-3}{4\sqrt{2}}-3\sqrt{\frac{1}{2}}\\\\=-3\sqrt{\frac{1}{2}}-\frac{3}{4\sqrt{2}}\\\\=-3\sqrt{\frac{1}{2}}-\frac{3\sqrt{2}}{8}$