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Ta có: \(\dfrac{-5}{13}\cdot\dfrac{3}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{7}\)
\(=\dfrac{5}{13}\left(-\dfrac{3}{7}+\dfrac{1}{7}\right)-\dfrac{2}{17}\cdot\dfrac{8}{13}\)
\(=\dfrac{5}{13}\cdot\dfrac{-2}{7}-\dfrac{2}{17}\cdot\dfrac{8}{13}\)
\(=\dfrac{-10}{91}-\dfrac{16}{221}\)
\(=\dfrac{-282}{1547}\)
Ta có: A = 1/1x2 + 1/2x3 + .... + 1/99x100
= 1x(1/1x2 + 1/2x3 + ... + 1/99x100)
= 1x(1 - 1/2 + 1/2 - 1/3 + ... + 1/99 - 1/100)
= 1x(1-1/100)
= 1 x 99/100
= 99/100
Ta có: 99/100 < 1 vì tử số bé hơn mẫu số
=> A = 1/1x2+1/2x3+...+1/99x100 < 1
(ko bik cách trình bày)
\(\dfrac{2}{3}\cdot x+50\%+x=\dfrac{5}{4}\)
\(\left(\dfrac{2}{3}+1\right)x=\dfrac{5}{4}-\dfrac{1}{2}\)
\(\dfrac{5}{3}x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{5}{3}\)
x=\(\dfrac{9}{20}\)
\(A=\frac{3}{4}-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(\frac{-1}{3}\right)^2\)
\(\Rightarrow A=\frac{3}{4}-\left(\frac{15}{6}+\frac{10}{6}\right)+\frac{1}{9}\)
\(\Rightarrow A=\frac{3}{4}-\frac{25}{6}+\frac{1}{9}\)
\(\Rightarrow A=\frac{18}{24}-\frac{100}{24}+\frac{1}{9}\)
\(\Rightarrow A=\frac{-82}{24}+\frac{1}{9}\)
\(\Rightarrow A=\frac{-41}{12}+\frac{1}{9}=\frac{-123}{36}+\frac{4}{36}=\frac{-119}{36}\)
Vậy A=\(\frac{-119}{36}\)
b)\(B=\frac{-11}{19}.\frac{4}{13}+\frac{-11}{13}.\frac{15}{19}+\frac{11}{13}\)
\(\Rightarrow B=\frac{-11}{19}.\frac{4}{13}+\frac{11}{13}.\left(\frac{-15}{19}+1\right)\)
\(\Rightarrow B=\frac{-11}{19}.\frac{4}{13}+\frac{11}{13}.\frac{4}{19}\)
\(\Rightarrow B=\frac{-11}{13}.\frac{4}{19}+\frac{11}{13}.\frac{4}{19}=\frac{4}{19}.\left(\frac{-11}{13}+\frac{11}{13}\right)=\frac{4}{19}.\frac{0}{13}=\frac{4}{19}.0=0\)
Vậy B=0
Chúc bn học tốt
(-12) . 13 + 13 . (-22)
= 13 . [-12 + (-22)]
= 13 . (-34)
= -442
(-12).13+13.(-22)
=13.(-12+-22)
=13.-34
=-442
\(\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{10}\right).\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}\)
\(=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}\)
\(=\dfrac{4.10.18...1558}{6.12.20...1560}\)
\(=\dfrac{41}{39}.3\)
\(=\dfrac{41}{11}\)
\(\dfrac{-3}{4}=\dfrac{9}{-12};\dfrac{-3}{9}=\dfrac{4}{-12};\dfrac{-12}{4}=\dfrac{9}{-3};\dfrac{-12}{9}=\dfrac{4}{-3}\)
\(B=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left(1-\dfrac{1}{10}\right)\cdot\left(1-\dfrac{1}{15}\right)\cdot...\cdot\left(1-\dfrac{1}{780}\right)\)
\(B=\left(1-\dfrac{1}{3+6+10+15+...+780}\right)\)
\(B=\left(1-\dfrac{1}{\left(780-3\right)\div3+1}\right)\)
\(B=\left(1-\dfrac{1}{260}\right)\)
\(B=\dfrac{259}{260}\)