Question

calculați:
a) 3/√2-√9/√32+√49/√8-√25/√2
c) √1,5-√3+3/8-5/6√1+1/2

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Answer 1

a)

$\frac{3}{ \sqrt{2} } - \frac{ \sqrt{9} }{ \sqrt{32} } + \frac{ \sqrt{49} }{ \sqrt{8} } - \frac{ \sqrt{25} }{ \sqrt{2} } = \frac{3}{ \sqrt{2} } - \frac{3}{4 \sqrt{2} } + \frac{7}{2 \sqrt{2} } - \frac{5}{ \sqrt{2} } =\frac{3 \times 4}{4 \sqrt{2} } - \frac{3}{4 \sqrt{2} } + \frac{7 \times 2}{4 \sqrt{2} } - \frac{5 \times 4}{ 4\sqrt{2} } = \frac{12 - 3 + 14 - 20}{4 \sqrt{2} } = \frac{3}{4 \sqrt{2} }$

c)

$\sqrt{1.5} - \sqrt{3 + \frac{3}{8} } - \frac{5}{6} \sqrt{1 + \frac{1}{2} } = \sqrt{ \frac{15}{10} } - \sqrt{ \frac{3 \times 8}{8} + \frac{3}{8} } - \frac{5}{6} \sqrt{ \frac{1 \times 2}{2} + \frac{1}{2} } = \frac{ \sqrt{3} }{ \sqrt{2} } - \sqrt{ \frac{24 + 3}{8} } - \frac{5}{6} \sqrt{ \frac{3}{2} } = \frac{ \sqrt{3} }{ \sqrt{2} } - \frac{ \sqrt{27} }{ \sqrt{8} } - \frac{5}{6} \frac{ \sqrt{3} }{ \sqrt{2} } = \frac{ \sqrt{3} }{ \sqrt{2} } (1 - \frac{5}{6}) - \frac{3 \sqrt{3} }{2 \sqrt{2} } = \frac{ 1}{6} \times \frac{ \sqrt{3} }{ \sqrt{2} } - \frac{3 \sqrt{3} }{2 \sqrt{2} } = \frac{ \sqrt{3} }{ \sqrt{2} } ( \frac{1 - 3}{6 - 2} ) = \frac{ - 2 \sqrt{3} }{4 \sqrt{2} }$