Question

17. Calculaţi modulul numărului real x în următoarele cazuri:
!DAU COROANA (clasa a VII-A)
Exercițiul 17, va rog mult!

Answer 1

a)

$x = \sqrt{0.18} - \sqrt{0.08} - \sqrt{0.72} = \sqrt{ \frac{18}{100} } - \sqrt{ \frac{8}{100} } - \sqrt{ \frac{72}{100} } = \frac{3 \sqrt{2} }{10} - \frac{2 \sqrt{2} }{10} - \frac{6 \sqrt{2} }{10} = \frac{ \sqrt{2}(3 - 2 - 6 )}{10} = \frac{ - 5 \sqrt{2} }{10} = \frac{ - \sqrt{2} }{2}$

$|x| = | \frac{ - \sqrt{2} }{2} | = \frac{ \sqrt{2} }{2}$

b)

$x = \sqrt{0.12} - \sqrt{0.27} + \sqrt{0.75} = \sqrt{ \frac{12}{100} } - \sqrt{ \frac{27}{100} } + \sqrt{ \frac{75}{100} } = \frac{2 \sqrt{3} }{10} - \frac{3 \sqrt{3} }{10} + \frac{5 \sqrt{3} }{10} = \frac{ \sqrt{3} (2 - 3 + 5)}{10} = \frac{4 \sqrt{3} }{10} = \frac{2 \sqrt{3} }{5}$

$|x| = \frac{2 \sqrt{3} }{5}$

c)

$x = \sqrt{0.45} - \sqrt{1.25} + \sqrt{2.45} = \sqrt{ \frac{45}{100} } - \sqrt{ \frac{125}{100} } + \sqrt{ \frac{245}{100} } = \frac{3 \sqrt{5} }{10} - \frac{5 \sqrt{5} }{10} + \frac{7 \sqrt{5} }{10} = \frac{ \sqrt{5}(3 - 5 + 7) }{10} = \frac{5 \sqrt{5} }{10} = \frac{ \sqrt{5} }{2}$

$|x| = \frac{ \sqrt{5} }{2}$

d)

$x = \sqrt{0.54} - \sqrt{0.24} - \sqrt{2.94} = \sqrt{ \frac{54}{100} } - \sqrt{ \frac{24}{100} } - \sqrt{ \frac{294}{100} } = \frac{9 \sqrt{6} }{10} - \frac{2 \sqrt{6} }{10} - \frac{7 \sqrt{6} }{10} = \frac{ \sqrt{6}(9 - 2 - 7) }{10} = 0$

$|x| = 0$