16 + 2 : 3 =
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a) \(\dfrac{-15+9+11}{16}=\dfrac{5}{16}\)
b) \(\dfrac{2}{3}\left(1,4+1,6-1,2\right)=\dfrac{2}{3}\times\dfrac{9}{5}=\dfrac{6}{5}\)
c) \(3\dfrac{2}{15}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{31}{15}=\dfrac{47}{15}-\dfrac{31}{15}=\dfrac{16}{15}\)
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
C = (2^16 - 2^8).(2^16 - 3^8).(2^16 - 4^8)
C=(216-28).(216-38).[(22)8-48]
C=(216-28).(216-38).(48-48)
C=(216-28).(216-38).0
C=0
\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ \Rightarrow A=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\\ \Rightarrow A=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{16}{2}+\dfrac{17}{2}\\ \Rightarrow A=\dfrac{1}{2}\left(2+3+4+...+17\right)=76\)
TL
= \(\frac{50}{3}\)
nhaaaaaa
HT