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a) Ta có A = 710 + 79 - 78
= 78( 72 + 7 - 1 )
= 78 . 55 ⋮ 11 vì 55 ⋮ 11
Vậy A ⋮ 11
b) Ta có B = 115 + 114 + 113
= 113( 112 + 11 + 1 )
= 113 . 133 ⋮ 7
Vậy B ⋮ 7
a,A=710+79-78=78(72+7-1)=78x55 ⋮11 vì 55⋮11
b,115+114+113=113(112+11+1)=113x133⋮7 vì 133⋮7
\(E=\left(-\dfrac{3}{7}+\dfrac{4}{11}\right):\dfrac{7}{11}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right):\dfrac{7}{11}\)
\(=\left(-\dfrac{3}{7}+\dfrac{4}{11}\right).\dfrac{11}{7}+\left(-\dfrac{4}{7}+\dfrac{7}{11}\right).\dfrac{11}{7}\)
\(=\dfrac{11}{7}\left[\left(-\dfrac{3}{7}\right)+\dfrac{4}{11}+\left(-\dfrac{4}{7}\right)+\dfrac{7}{11}\right]\)
\(=\dfrac{11}{7}\left[\left(-\dfrac{3}{7}+\dfrac{-4}{7}\right)+\left(\dfrac{4}{11}+\dfrac{7}{11}\right)\right]\)
\(=\dfrac{11}{7}\left[\left(-1\right)+1\right]\)
\(=\dfrac{11}{7}.0=0\)
\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}-\dfrac{9}{11}-\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\left(+\dfrac{7}{9}\rightarrow-\dfrac{7}{9}\right)\)
\(\Rightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}\)
\(\Rightarrow A=-\dfrac{11}{13}+\dfrac{13}{15}\)
\(\Rightarrow A=\dfrac{-11.15+13.13}{13.15}\)
\(\Rightarrow A=\dfrac{-165+169}{195}=\dfrac{4}{195}\)
a: =11/7(-3/7+4/11-4/7+7/11)=0
b: \(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{95}-\dfrac{1}{97}\right)\)
\(=\dfrac{1}{99\cdot97}-\dfrac{1}{2}\cdot\dfrac{96}{97}=\dfrac{1}{99\cdot97}-\dfrac{48}{97}=-\dfrac{4751}{9603}\)
=-17/44:7/11+5/77:7/11
=(-1 7/44+5/77):7/11
=-9/28:7/11
=-99/196
\(A=\dfrac{7}{38}\left(\dfrac{9}{11}+\dfrac{4}{11}-\dfrac{2}{11}\right)=\dfrac{7}{38}\cdot1=\dfrac{7}{38}\)