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\(B=2+2^2+2^3+2^4+...+2^{99}+2^{100}=2\left(1+2^2+2^3+2^4\right)+...+2^{96}\left(1+2^2+2^3+2^4\right)=2.31+2^6.31+...+2^{96}.31=31\left(2+2^6+...+2^{96}\right)⋮31\)
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\(a.120=2^3.3.5\)
\(320=2^6.5\)
\(ƯC\left(120;320\right)=\left\{40;80;120;160;200;...\right\}\)
\(ƯCLN\left(120;320\right)=2^3.5=40\)
\(b.455=5.7.13\)
\(950=2.5^2.19\)
\(ƯC\left(455;950\right)=\left\{5;10;15;20;25;...\right\}\)
\(ƯCLN\left(455;950\right)=5\)
\(c.126=2.3^2.7\)
\(108=2^2.3^3\)
\(306=2.3^2.17\)
\(ƯC\left(126;108;306\right)=\left\{18;36;54;72;...\right\}\)
\(ƯCLN\left(126;108;306\right)=2.3^2=18\)
a.120=23.3.5�.120=23.3.5
320=26.5320=26.5
ƯC(120;320)={40;80;120;160;200;...}Ư�(120;320)={40;80;120;160;200;...}
ƯCLN(120;320)=23.5=40Ư���(120;320)=23.5=40
b.455=5.7.13�.455=5.7.13
950=2.52.19950=2.52.19
ƯC(455;950)={5;10;15;20;25;...}Ư�(455;950)={5;10;15;20;25;...}
ƯCLN(455;950)=5Ư���(455;950)=5
c.126=2.32.7�.126=2.32.7
108=22.33108=22.33
306=2.32.17306=2.32.17
ƯC(126;108;306)={18;36;54;72;...}Ư�(126;108;306)={18;36;54;72;...}
ƯCLN(126;108;306)=2.32=18
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\(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{13.15}\right).x=\dfrac{-26}{45}\\ \Leftrightarrow\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{13.15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{13}-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\left(1-\dfrac{1}{15}\right).x=\dfrac{-52}{45}\\ \Leftrightarrow\dfrac{14}{15}.x=\dfrac{-52}{45}\\ \Leftrightarrow x=-\dfrac{26}{21}\)
(11.3+13.5+...+113.15).x=−2645⇔(21.3+23.5+...+213.15).x=−5245⇔(1−13+13−15+...+113−115).x=−5245⇔(1−115).x=−5245⇔1415.x=−5245⇔x=−2621
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