Explicație pas cu pas:
36. a)
[tex]- 5 \cdot \{2 + \left[ {( - 150)}^{3} \div {( - 75)}^{3} + {( - 5)}^{6} \times {4}^{6} \div {( - 20)}^{6} \right] \div ( - 3)\} = \\ [/tex]
[tex] = - 5 \cdot \{2 + \left[ {(2 \times 75)}^{3} \div {75}^{3} + {5}^{6} \times {4}^{6} \div {20}^{6} \right] \div ( - 3)\} \\ [/tex]
[tex]= - 5 \cdot \{2 + \left[ {2}^{3} \times {75}^{3} \div {75}^{3} + ({5} \times {4})^{6} \div {20}^{6} \right] \div ( - 3)\} \\ [/tex]
[tex]= - 5 \cdot \left[2 + ({2}^{3} + 20^{6} \div {20}^{6})\div ( - 3) \right] \\ [/tex]
[tex]= - 5 \cdot \left[2 + (8 + 1)\div ( - 3) \right][/tex]
[tex]= - 5 \cdot \left[2 + 9\div ( - 3) \right][/tex]
[tex]= - 5 \cdot (2 - 3)[/tex]
[tex]= - 5 \cdot ( - 1) = 5[/tex]
b)
[tex]{( - 5)}^{30} \div \{ - 1 + 3 \cdot \left[ {( - 5)}^{20} \div {25}^{10} + {( - 22 + 17)}^{15} \div ( - {125}^{5} ) \right] \} = \\ [/tex]
[tex]= {5}^{30} \div \{ - 1 + 3 \cdot \left[ {5}^{20} \div {( {5}^{2} )}^{10} + {( - 5)}^{15} \div ( - {( {5}^{3} )}^{5} ) \right] \} \\ [/tex]
[tex]= {5}^{30} \div \{ - 1 + 3 \cdot \left[ {5}^{20} \div {5}^{20} + {( - 5}^{15}) \div ( - {5}^{15} ) \right] \} \\ [/tex]
[tex]= {5}^{30} \div \left[- 1 + 3 \cdot (1 + 1) \right] \\ [/tex]
[tex]= {5}^{30} \div (- 1 + 3 \cdot 2)[/tex]
[tex]= {5}^{30} \div (- 1 + 6)[/tex]
[tex]= {5}^{30} \div 5 = {5}^{29}[/tex]
37.
[tex]a = (2 - 2 \cdot 5) \cdot \left[ {( - 7)}^{42} \div {49}^{20} + 25 \div ( - {5}^{2} ) + 7 \cdot ( - {2}^{2} ) \right] \div 40 = \\ [/tex]
[tex]= (2 - 10) \cdot \left[ {7}^{42} \div {( {7}^{2} )}^{20} + 25 \div ( - 25) + 7 \cdot ( - 4) \right] \div 40 \\ [/tex]
[tex]= ( - 8) \cdot ({7}^{42} \div {7}^{40} - 1 - 28) \div 40 \\ [/tex]
[tex]= ( - 8) \cdot ({7}^{2} - 29) \div 40[/tex]
[tex]= ( - 8) \cdot (49 - 29) \div 40[/tex]
[tex]= ( - 8) \cdot 20 \div 40 [/tex]
[tex]= - 160 \div 40 = - 40[/tex]
[tex]\implies a = - 40[/tex]
.[tex]b = | - {3}^{2} | - { | - 3| }^{2} + \left[ - 12 - | - 5| + {6}^{4} \div {( - 3)}^{4} \right] + {32}^{12} \div {( - 4)}^{30} = \\ [/tex]
[tex]= {3}^{2} - {3}^{2} + \left[ - 12 - 5 + {(2 \times 3)}^{4} \div {3}^{4} \right] + {( {2}^{5} )}^{12} \div {( {2}^{2} )}^{30} \\ [/tex]
[tex]= 0 + (- 17 + {2}^{4} \times {3}^{4} \div {3}^{4}) + {2}^{60} \div {2}^{60} \\ [/tex]
[tex]= (- 17 + {2}^{4}) + 1[/tex]
[tex] = - 17 + 16 + 1 = 0[/tex]
[tex]\implies b = 0[/tex]
→
[tex]a < b[/tex]
