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a, \(A=2015.20162016-2016.20152015\)
\(A=2015.\left(2016.10001\right)-2016.20152015\)
\(A=\left(2015.10001\right).2016-20152015.2016\)
\(A=20152015.2016-20152015.2016\)
\(A=0\)
Vậy A = 0
b, \(B=\left(3.4.2^{16}\right)^2\div11.2^{13}.4^{11}-16^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9\)
\(B=3^2.2^4.2^{32}\div11.2^{13}.2^{22}-2^{36}\)
\(B=3^2.2^{36}\div11.2^{35}-2^{36}\)
\(B=3^2.2^{35}.2\div11.2^{35}-2.2^{35}\)
\(B=3^2.2\div9=9.2\div9=2\)
Vậy B = 2
c, \(C=2^{10}.13+2^{10}.65\div2^8.104\)
\(C=2^{10}.\left(13+65\right)\div2^8.104\)
\(C=2^{10}.78\div2^8.104\)
\(C=2^{10}.39\div2^8.13\)
\(C=39\div13=3\)
Vậy C = 3
Đề bài câu c sai mk sửa nhé là 28 ms tính đc k nó dư lắm !!!
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\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}.\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=3\)
\(A=\frac{2^{10}.13+2^{10}.65}{2^8.104}=\frac{2^{10}\left(13+65\right)}{2^8.104}=\frac{2^{10}.78}{2^8.104}=\frac{2^2.78}{104}=\frac{2^2.2.39}{2^3.13}=\frac{2^3.39}{2^3.13}=3\)
A=\(\frac{72^3.54^2}{108^4}=\frac{\left(2^3.3^2\right)^3.\left(2.3^3\right)^2}{\left(2^2.3^3\right)^4}=\frac{2^9.3^6.2^2.3^6}{2^8.3^{12}}=\frac{2^{11}.3^{12}}{2^8.3^{12}}=2^3=8\)
B= \(\frac{4^6.3^4.9^5}{6^{12}}=\frac{2^{12}.3^4.3^{10}}{2^{12}.3^{12}}=\frac{2^{12}.3^{14}}{2^{12}.3^{12}}=3^2=9\)
c) \(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\left(2^8+1\right)}{2^2\left(2^8+1\right)}=2^3=8\)
1.
\(\frac{72^3\times54^2}{108^4}=\frac{\left(8\times9\right)^3\times\left(27\times2\right)^2}{\left(27\times4\right)^4}=\frac{\left(2^3\times3^2\right)^3\times\left(3^3\times2\right)^2}{\left(3^3\times2^2\right)^4}=\frac{\left(2^3\right)^3\times\left(3^2\right)^3\times\left(3^3\right)^2\times2^2}{\left(3^3\right)^4\times\left(2^2\right)^4}=\frac{2^9\times3^6\times3^6\times2^2}{3^{12}\times2^8}=2^3=8\)
2.
\(\frac{4^6\times3^4\times9^5}{6^{12}}=\frac{\left(2^2\right)^6\times3^4\times\left(3^2\right)^5}{\left(2\times3\right)^{12}}=\frac{2^{12}\times3^4\times3^{10}}{2^{12}\times3^{12}}=3^2=9\)
3.
\(\frac{2^{13}+2^5}{2^{10}+2^2}=\frac{2^5\times\left(2^8+1\right)}{2^2\times\left(2^8+1\right)}=2^3=8\)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{10}{33}\)
\(F=\frac{10}{99}\)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
mình hỏi xíu là 101+105 ở đâu zị :)?