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\(\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(2^4-4^2\right)\)
\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).\left(16-16\right)\)
\(=\left(2^{17}+7^2\right).\left(9^{15}-3^{15}\right).0\)
\(=0\)
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\(\left(7^{1997}-7^{1995}\right):\left(7^{1994}.7\right)\)
\(=\left[7^{1995}\left(7^2-1\right)\right]:7^{1995}\)
\(=7^{1995}.48:7^{1995}\)
\(=48\)
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\(\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(3^8-81^2\right)\)
\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).\left(6561-6561\right)\)
\(=\left(1^2+2^3+3^4+4^5\right).\left(1^3+2^3+3^3+4^3\right).0\)
\(=0\)
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\(\left(2^8+8^3\right):\left(2^5.2^3\right)\)
\(=\left[2^8+\left(2^3\right)^3\right]:2^8\)
\(=\left(2^8+2^9\right):2^8\)
\(=2^8.\left(1+2\right):2^8\)
\(=2^8.3:2^8\)
\(=3\)
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3A=1.2.3+2.3.(4-1)+3.4.(5-2)+............+2015.2016.(2017-2014)
3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+.............+2015.2016.2017-2014.2015.2016
3A=2015.2016.2017
A=2015.2016.2017:3
A=2015.672.2017
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a.\(\frac{5100-10}{8160-16}\)= \(\frac{5090}{8144}\) = \(\frac{5}{8}\)
b.\(\frac{952+8}{2142+18}\) = \(\frac{960}{2160}\) = \(\frac{4}{9}\)
Chúc bạn học tốt!
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2A=2+22+23+24+...+211
2A—A=(2+22+23+24+....+211)—(1+2+22+23+...+210)
A=211—1
Ta có A = 2A - A
= \(2\left(1+2+2^2+2^3+...+2^{10}\right)\)- \(\left(1+2+2^2+2^3+....+2^{10}\right)\)
=\(2+2^2+2^3+2^4+.....+2^{11}\)\(-1-2-2^2-2^3-...-2^{10}\)
=\(2^{11}-1\)(Các số còn lại đã trừ hết cho nhau)
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P = 8.( 7 - 72 + 73 - 74 +...+ 72022)
Đặt B = 7 - 72 + 73 - 74+...+ 72022
7 \(\times\)B = 72 - 73 + 74-....- 72022 + 72023
7B + B = 7 + 72023
8B = ( 7 + 72023)
B = ( 7 + 72023): 8
P = 8 \(\times\) ( 7 + 72023) : 8
P = 7 + 72023