Question

poate cineva va rog sa ma ajute? nu inteleg nimic​

Answer 1

Explicație pas cu pas:

primul modul:

$\dfrac{1}{ {8}^{35} } + \dfrac{1}{ {( - 5)}^{45} } = \dfrac{1}{ {( {2}^{3} )}^{35} } + \dfrac{1}{ - {5}^{45} } = \dfrac{1}{ {2}^{3 \cdot 35} } - \dfrac{1}{{5}^{45} } = \dfrac{{5}^{45} - {2}^{105}}{ {2}^{105} \cdot {5}^{45}}$

${5}^{45} - {2}^{105} = {({5}^{3})}^{15} - {({2}^{7})}^{15} = {125}^{15} - {128}^{15} < 0$

$\implies \dfrac{{5}^{45} - {2}^{105}}{ {2}^{105} \cdot {5}^{45}} < 0$

$\implies \bigg|\dfrac{{5}^{45} - {2}^{105}}{ {2}^{105} \cdot {5}^{45}}\bigg| = \dfrac{{2}^{105} - {5}^{45}}{ {2}^{105} \cdot {5}^{45}}$

al doilea modul:

$\dfrac{1}{ {2}^{105} } + \dfrac{1}{ {( - 25)}^{21} } = \dfrac{1}{{2}^{105}} + \dfrac{1}{ - {25}^{21} } = \dfrac{1}{{2}^{105}} - \dfrac{1}{{( {5}^{2} )}^{21} } = \dfrac{1}{{2}^{105}} - \dfrac{1}{{5}^{42}} = \dfrac{{5}^{42} - {2}^{105}}{ {2}^{105} \cdot {5}^{42}}$

${5}^{42} - {2}^{105} = {({5}^{2})}^{21} - {({2}^{5})}^{21} = {25}^{21} - {32}^{21} < 0$

$\implies \dfrac{{5}^{42} - {2}^{105}}{ {2}^{105} \cdot {5}^{42}} < 0$

$\implies \bigg|\dfrac{{5}^{42} - {2}^{105}}{ {2}^{105} \cdot {5}^{42}}\bigg| = \dfrac{{2}^{105} - {5}^{42}}{ {2}^{105} \cdot {5}^{42}}$

atunci:

$= \bigg(\dfrac{{2}^{105} - {5}^{45}}{ {2}^{105} \cdot {5}^{45}} + \dfrac{1}{{( {2}^{2} )}^{52}} + \dfrac{{2}^{105} - {5}^{42}}{ {2}^{105} \cdot {5}^{42}}\bigg) : \dfrac{14}{{5}^{43}}$

$= \bigg(\dfrac{{2}^{105} - {5}^{45}}{ {2}^{105} \cdot {5}^{45}} + \dfrac{1}{{2}^{104}} + \dfrac{{2}^{105} - {5}^{42}}{ {2}^{105} \cdot {5}^{42}}\bigg) \cdot \dfrac{{5}^{43}}{14}$

$= \bigg(\dfrac{{2}^{105} - {5}^{45}}{ {2}^{105} \cdot {5}^{45}} + \dfrac{2 \cdot {5}^{45}}{{2}^{105} \cdot {5}^{45}} + \dfrac{{5}^{3} \cdot ({2}^{105} - {5}^{42})}{ {2}^{105} \cdot {5}^{45}}\bigg) \cdot \dfrac{{5}^{43}}{14}$

$= \dfrac{{2}^{105} - {5}^{45} + 2 \cdot {5}^{45} + {5}^{3} \cdot {2}^{105} - {5}^{45}}{ {2}^{105} \cdot {5}^{45}} \cdot \dfrac{{5}^{43}}{14}$

$= \dfrac{({2}^{105} + 125 \cdot {2}^{105}) + (2 \cdot {5}^{45} - 2{5}^{45})}{ {2}^{105} \cdot {5}^{45 - 43}} \cdot \dfrac{1}{14}$

$= \dfrac{ 126 \cdot {2}^{105}}{ {2}^{105} \cdot {5}^{2}} \cdot \dfrac{1}{14} = \bf \dfrac{9 }{25}$