Question

ajutor va rog la ex 13, 14 și 15​

Answer 1

Explicație pas cu pas:

13. a.

$\frac{ \sqrt{144} - \sqrt{81} }{x} = \frac{ \sqrt{225} }{ \sqrt{900} + \sqrt{25} } \\ \frac{12 - 9}{x} = \frac{15}{30 + 5} \\ \frac{3}{x} = \frac{15}{35} \\ \frac{3}{x} = \frac{3}{7} = > x = 7$

13. b.

$\frac{ \sqrt{32 + \sqrt{1024} } }{ \sqrt{135 - \sqrt{1225} } } = \frac{2x}{ \sqrt{270 - \sqrt{2025} } } \\ \frac{ \sqrt{32 + 32} }{ \sqrt{135 - 35} } = \frac{2x}{ \sqrt{270 - 45} } \\ \frac{ \sqrt{64} }{ \sqrt{100} } = \frac{2x}{ \sqrt{225} } < = > \frac{8}{10} = \frac{2x}{15} \\ x = \frac{8 \times 15}{2 \times 10} = > x = 6$

14.

$x = 25 \times \left[ \sqrt{4 \times 0.(4)} + \sqrt{40 \times 0.0(4)} + \sqrt{400 \times 0.00(4)}\right] \\ = 25 \times\left(\sqrt{4 \times \frac{4}{9} } + \sqrt{40 \times \frac{4}{90} } + \sqrt{400 \times \frac{4}{900} } \right)$

$= 25 \times \left(\sqrt{\frac{ {4}^{2} }{ {3}^{2} }} + \sqrt{\frac{ {4}^{2} }{ {3}^{2} } } + \sqrt{\frac{ {4}^{2} }{ {3}^{2} } } \right)$

$= 25 \times \left(3 \times \frac{4}{3} \right) = 25 \times 4 = 100 = {10}^{2}$

15.

$\frac{1}{1 \times 2} + \frac{1}{2 \times 3} +...+ \frac{1}{24 \times 25} = \\ = \frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{24} - \frac{1}{25} \\ = \frac{1}{1} - \frac{1}{25} = \frac{25 - 1}{25} = \frac{24}{25}$

$\frac{1}{25 \times 26} + \frac{1}{26 \times 27} + ... + \frac{1}{49 \times 50} = \\ = \frac{1}{25} - \frac{1}{26} + \frac{1}{26} - \frac{1}{27} + ... + \frac{1}{49} - \frac{1}{50} \\ = \frac{1}{25} - \frac{1}{50} = \frac{2 - 1}{50} = \frac{1}{50}$

$x = \sqrt{ \frac{1}{94}\cdot \left( \frac{24}{25} - \frac{1}{50} \right)} = \sqrt{ \frac{1}{94}\cdot \frac{48 - 1}{50} } \\ = \sqrt{ \frac{1}{94}\cdot \frac{47}{50}} = \sqrt{ \frac{1}{2\cdot 50} } = \sqrt{ \frac{1}{100} } = \frac{1}{10}$