Question

Ex 20 va eog dau 100 puncte urgent

Answer 1

Răspuns:

b) 2; c) 2

Explicație pas cu pas:

b)

$\dfrac{3}{1\cdot 4} + \dfrac{3}{4 \cdot 7} + \dfrac{3}{7 \cdot 10} + ... + \dfrac{3}{58 \cdot 61} + \dfrac{3}{61 \cdot 64} = \dfrac{4 - 1}{1\cdot 4} + \dfrac{7 - 4}{4 \cdot 7} + \dfrac{10 - 7}{7 \cdot 10} + ... + \dfrac{61 - 58}{58 \cdot 61} + \dfrac{64 - 61}{61 \cdot 64} = \dfrac{1}{1} -\not \dfrac{1}{4} +\not \dfrac{1}{4} -\not \dfrac{1}{7} +\not \dfrac{1}{7} -\not \dfrac{1}{10} + ... +\not \dfrac{1}{58} -\not \dfrac{1}{61} +\not \dfrac{1}{61} - \dfrac{1}{64} = \dfrac{1}{1} - \dfrac{1}{64} = \dfrac{64 - 1}{64} =\bf \dfrac{63}{64}$

și

$\dfrac{2}{1\cdot 3} + \dfrac{2}{3 \cdot 5} + \dfrac{2}{5 \cdot 7} + ... + \dfrac{2}{59 \cdot 61} + \dfrac{2}{61 \cdot 63} = \dfrac{3 - 1}{1\cdot 3} + \dfrac{5 - 3}{3 \cdot 5} + \dfrac{7 - 5}{5 \cdot 7} + ... + \dfrac{61 - 59}{59 \cdot 61} + \dfrac{63 - 61}{61 \cdot 63} = \dfrac{1}{1} -\not \dfrac{1}{3} +\not \dfrac{1}{3} -\not \dfrac{1}{5} +\not \dfrac{1}{5} -\not \dfrac{1}{7} + ... +\not \dfrac{1}{59} -\not \dfrac{1}{61} +\not \dfrac{1}{61} - \dfrac{1}{63} = \dfrac{1}{1} - \dfrac{1}{63} = \dfrac{63 - 1}{63} = \bf \dfrac{62}{63}$

=>

$\sqrt{128 \cdot 126 \cdot \Big( \dfrac{63}{64} - \dfrac{62}{63}\Big)} = \sqrt{128 \cdot 126 \cdot \dfrac{63 \cdot 63 - 62 \cdot 64}{64 \cdot 63}}$

$= \sqrt{2 \cdot 64 \cdot 2 \cdot 63 \cdot \dfrac{1}{64 \cdot 63}} = \sqrt{ {2}^{2} } = \bf 2$

c)

$\dfrac{1}{1\cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + ... + \dfrac{1}{43 \cdot 44} + \dfrac{1}{44 \cdot 45} = \dfrac{2 - 1}{1\cdot 2} + \dfrac{3 - 2}{2 \cdot 3} + \dfrac{4 - 3}{3 \cdot 4} + ... + \dfrac{44 - 43}{43 \cdot 44} + \dfrac{45 - 44}{44 \cdot 45} = \dfrac{1}{1} - \not\dfrac{1}{2} + \not\dfrac{1}{2} - \not\dfrac{1}{3} + \not\dfrac{1}{3} - \not\dfrac{1}{4} + ... + \not\dfrac{1}{43} - \not\dfrac{1}{44} + \not\dfrac{1}{44} - \dfrac{1}{45} = \dfrac{1}{1} - \dfrac{1}{45} = \dfrac{45 - 1}{45} = \bf \dfrac{44}{45}$

și

$\dfrac{1}{45\cdot 46} + \dfrac{1}{46 \cdot 47} + \dfrac{1}{47 \cdot 48} + ... + \dfrac{1}{88 \cdot 89} + \dfrac{1}{89 \cdot 90} = \dfrac{46 - 45}{45\cdot 46} + \dfrac{47 - 46}{46 \cdot 47} + \dfrac{48 - 47}{47 \cdot 48} + ... + \dfrac{89 - 88}{88 \cdot 89} + \dfrac{90 - 89}{89 \cdot 90} = \dfrac{1}{45} - \dfrac{1}{46} + \dfrac{1}{46} - \dfrac{1}{47} + \dfrac{1}{47} - \dfrac{1}{48} + ... + \dfrac{1}{88} - \dfrac{1}{89} + \dfrac{1}{89} - \dfrac{1}{90} = \dfrac{1}{45} - \dfrac{1}{90} = \dfrac{2 - 1}{90} = \bf \dfrac{1}{90}$

=>

$\sqrt{ \dfrac{120}{29} \cdot \Big( \dfrac{44}{45} - \dfrac{1}{90}\Big)} = \sqrt{ \dfrac{120}{29} \cdot \dfrac{88 - 1}{90}}$

$= \sqrt{ \dfrac{120}{29} \cdot \dfrac{87}{90}} = \sqrt{ \dfrac{4}{1} \cdot \dfrac{3}{3}} = \sqrt{4} = \bf 2$