Tính:
a) (6² + 7² + 8² + 9² + 10²) – (1² + 2² + 3² + 4² + 5²)
b) (125³ . 7^5 – 175^5 : 5) : 2001^2002
c)16 . 64 . 8^2 : (4^3 . 2^5 . 16)
Tính theo trình tự nha
Tính:
a) (6² + 7² + 8² + 9² + 10²) – (1² + 2² + 3² + 4² + 5²)
b) (125³ . 7^5 – 175^5 : 5) : 2001^2002
c)16 . 64 . 8^2 : (4^3 . 2^5 . 16)
Tính theo trình tự nha
Đáp án:
$\begin{array}{l}
a)\left( {{6^2} + {7^2} + {8^2} + {9^2} + {{10}^2}} \right)\\
– \left( {{1^2} + {2^2} + {3^2} + {4^2} + {5^2}} \right)\\
= \left( {{6^2} – {1^2}} \right) + \left( {{7^2} – {2^2}} \right)\\
+ \left( {{8^2} – {3^2}} \right) + \left( {{9^2} – {4^2}} \right) + \left( {{{10}^2} – {5^2}} \right)\\
= \left( {6 – 1} \right)\left( {6 + 1} \right) + \left( {7 – 2} \right)\left( {7 + 2} \right)\\
+ \left( {8 – 3} \right)\left( {8 + 3} \right) + \left( {9 – 4} \right)\left( {9 + 4} \right) + \left( {10 – 5} \right)\left( {10 + 5} \right)\\
= 5.\left( {1 + 6} \right) + 5.\left( {2 + 7} \right) + 5.\left( {3 + 8} \right) + 5.\left( {4 + 9} \right) + 5.\left( {5 + 10} \right)\\
= 5.\left( {1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10} \right)\\
= 5.\frac{{\left( {10 + 1} \right).10}}{2}\\
= 275\\
b)\left( {{{125}^3}{{.7}^5} – {{175}^5}:5} \right):{2001^{2002}}\\
= \left( {{5^{3.3}}{{.7}^5} – {{\left( {{{7.5}^2}} \right)}^5}:5} \right):{2001^{2002}}\\
= \left( {{5^9}{{.7}^5} – {7^5}{{.5}^{10}}:5} \right):{2001^{2002}}\\
= \left( {{5^9}{{.7}^5} – {5^9}{{.7}^5}} \right):{2001^{2002}}\\
= 0:{2001^{2002}}\\
= 0\\
c){16.64.8^2}:\left( {{4^3}{{.2}^5}.16} \right)\\
= {2^4}{.2^6}{.2^6}:\left( {{2^6}{{.2}^5}{{.2}^4}} \right)\\
= {2^{16}}:{2^{15}}\\
= 2
\end{array}$