Explicație pas cu pas:
d)
[tex]( - 2) \cdot {( - 2)}^{2}\cdot {( - 2)}^{3}\cdot ...\cdot {( - 2)}^{20} \div {( - 8)}^{70} = \\ [/tex]
[tex]= ({ - 2})^{1 + 2 + 3 + ... + 20} \div {8}^{70} \\ [/tex]
[tex]= {( - 2)}^{ \frac{20 \times 21}{2} } \div {( {2}^{3} )}^{70} \\ [/tex]
[tex]= {( - 2)}^{210} \div {2}^{210} \\ [/tex]
[tex]= {2}^{210} \div {2}^{210} = 1[/tex]
a)
[tex]( | {2}^{30} - {3}^{30} | + {2}^{30} ) \div {( - 27)}^{10} = ({3}^{30} - {2}^{30} + {2}^{30}) \div {27}^{10} = \\ [/tex]
[tex]= {3}^{30} \div {( {3}^{3} )}^{10} = {3}^{30} \div {3}^{30} = 1[/tex]
b)
[tex]\left[ | {3}^{12} - {5}^{8} | + {( - 25)}^{4} \right] \cdot {( - 2)}^{12} - {36}^{6} = \\ [/tex]
[tex]= \left[ | {( {3}^{3} )}^{4} - {( {5}^{2} )}^{4} | + {25}^{4} \right] \cdot {2}^{12} - {({6}^{2})}^{6} \\ [/tex]
[tex]= (| {27}^{4} - {25}^{4} | + {25}^{4}) \cdot {2}^{12} - {6}^{12} \\ [/tex]
[tex]= ({27}^{4} - {25}^{4} + {25}^{4}) \cdot {2}^{12} - {6}^{12} \\ [/tex]
[tex]= {3}^{12} \cdot {2}^{12} - {6}^{12} = (3 \cdot 2)^{12} - {6}^{12} \\ [/tex]
[tex]= {6}^{12} - {6}^{12} = 0[/tex]
c)
[tex]{4}^{15} - 3 \cdot {4}^{14} - 3 \cdot {4}^{13} - ... - 3 \cdot {4} = \\ [/tex]
[tex]= {4}^{15} - 3({4}^{14} + {4}^{13} + ... + {4}^{1}) \\ [/tex]
[tex]= {4}^{15} - 3 \cdot \frac{4( {4}^{14} - 1)}{4 - 1} = {4}^{15} - 3 \cdot \frac{{4}^{15} - 4}{3} \\ [/tex]
[tex]= {4}^{15} - ({4}^{15} - 4) = {4}^{15} - {4}^{15} + 4 = 4 \\ [/tex]
d)
[tex]{5}^{100} - 4 \cdot {5}^{99} - 4 \cdot {5}^{98} - ... - 4 \cdot {5} = \\ [/tex]
[tex]= {5}^{100} - 4({5}^{99} + {5}^{98} + ... + {5}^{1}) \\ [/tex]
[tex]= {5}^{100} - 4 \cdot \frac{5( {5}^{99} - 1)}{5 - 1} = {5}^{100} - 4 \cdot \frac{{5}^{100} - 5}{4} \\ [/tex]
[tex]= {5}^{100} - ({5}^{100} - 5) = {5}^{100} - {5}^{100} + 5 = 5 \\ [/tex]
.
a)
[tex]( - 5 - 10 - 15 - ... - 400) \div ( - 1 - 2 - 3 - ... - 80) = \\ [/tex]
[tex]= \left[( - 5)(1 + 2 + 3 + ... + 80) \right] \div \left[( - 1)(1 + 2 + 3 + ... + 80) \right] \\ [/tex]
[tex]= ( - 5) \div ( - 1) = 5[/tex]
