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\(A=\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)
Ta có :
\(1+3^4+3^8+3^{12}=\left(3^{16}-1\right):\left(3^4-1\right)\)
\(1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}=\left(3^{16}-1\right):\left(3^2-1\right)\)
\(\Rightarrow A=\frac{\left(3^{16}-1\right):\left(3^4-1\right)}{\left(3^{16}-1\right):\left(3^2-1\right)}=\frac{3^2-1}{3^4-1}=\frac{1}{10}\)
Ta có
• A=1+34+38+312
=>34.A=34+38+312+316
<=>81.A-A=316-1
<=>A=(316-1)/80=538084
•B=1+32+34+36+38+310+312+314
=>32.B=32+34+36+38+310+312+314+316
<=>8.B=316-1
<=>B=(316-1)/8=53808400
Vậy Q=A/B=538084/53808400=1/100=0.01
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
\(A=\dfrac{4^{10}+8^4}{4^5+8^6}\)
\(A=\dfrac{2^{20}+2^{12}}{2^{10}+2^{18}}=\dfrac{\left(2^8+1\right).2^{12}}{\left(1+2^8\right).2^{10}}\)
\(=\dfrac{\left(256+1\right).2^2}{1+256}=\dfrac{257.2^2}{257}=2^2\)
\(B=\dfrac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)
\(=\dfrac{1+81+6561+3^{12}}{1+9+81+729+6561+59049+3^{12}+3^{14}}\)
\(=\dfrac{6643+3^{12}}{91+719+6561+59049+3^{12}+3^{14}}\)
\(=\dfrac{6643+3^{12}}{66430+3^{12}+3^{14}}\)
P/s : Nổi hứng lên thì lm chứ k bt đúng hay sai :V
\(B=\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)
\(B=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+3^2+3^6+3^{10}+3^{14}}\)
Xét mẫu : \(\left(1+3^4+3^8+3^{12}\right)+3^2\left(1+3^4+3^8+3^{12}\right)=\left(3^2+1\right)\left(1+3^4+3^8+3^{12}\right)\)
Ta có : \(\frac{1+3^4+3^8+3^{12}}{\left(3^2+1\right)\left(1+3^4+3^8+3^{12}\right)}=\frac{1}{3^2+1}=\frac{1}{10}\)
(chủ câu hỏi)
sao lại có \(3^2+1\)ở kia