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\(B=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...\dfrac{1}{200}\right)>\dfrac{1}{150}+..\dfrac{1}{150}+\dfrac{1}{200}+..+200=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)Vậy ... (ta có điều phải chứng minh )
Ta có :\(\dfrac{1}{20}>\dfrac{1}{200}\)
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\(\dfrac{1}{199}>\dfrac{1}{200}\)
Do đó : \(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+..+\dfrac{1}{200}=\dfrac{181}{200}>\dfrac{180}{200}=\dfrac{9}{10}\)Vậy ...
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
\(\dfrac{5}{12}+\dfrac{-7}{12}=\dfrac{-2}{12}=\dfrac{-1}{6}\)
\(\dfrac{1}{2}+\dfrac{-2}{3}=\dfrac{3}{6}+\dfrac{-4}{6}=\dfrac{-1}{6}\)
\(\dfrac{3}{5}-\dfrac{4}{3}=\dfrac{9}{15}-\dfrac{20}{15}=\dfrac{-11}{15}\)
\(\dfrac{-15}{14}.\dfrac{21}{20}=\dfrac{-315}{280}=\dfrac{-9}{8}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-1}{6}\)
\(\dfrac{-11}{15}\)
\(\dfrac{-9}{8}\)
\(\dfrac{5}{7}-x=\dfrac{9}{21}\\ \Rightarrow\dfrac{5}{7}-x=\dfrac{3}{7}\\ \Rightarrow x=\dfrac{5}{7}-\dfrac{3}{7}\\ \Rightarrow x=\dfrac{2}{7}\\ b,-x-\dfrac{1}{3}=\dfrac{2}{6}\\ \Rightarrow-x=\dfrac{2}{6}+\dfrac{1}{3}\\ \Rightarrow-x=\dfrac{2}{3}\\ \Rightarrow x=-\dfrac{2}{3}\\ \dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\\ \Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\\ \Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\\ \Rightarrow x=-\dfrac{13}{12}\)
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
