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\(\dfrac{1}{2}N=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
N=\(\dfrac{2}{5}:\dfrac{1}{2}=\dfrac{4}{5}\)
#)Trả lời :
\(A=\frac{\left(140+70+42+28+20+15\right)}{420}\)
\(A=\frac{315}{420}=\frac{\left(315:105\right)}{\left(420:105\right)}=\frac{3}{4}\)
Vậy : \(A=\frac{3}{4}\)
#~Will~be~Pens~#
a: \(=\dfrac{1}{3}-\dfrac{6}{13}\cdot\dfrac{13}{36}\cdot\dfrac{9}{10}\)
\(=\dfrac{1}{3}-\dfrac{1}{6}\cdot\dfrac{9}{10}\)
\(=\dfrac{1}{3}-\dfrac{3}{20}=\dfrac{20-9}{60}=\dfrac{11}{60}\)
b: \(=\dfrac{1}{3}\cdot\dfrac{15}{7}\cdot\dfrac{34}{45}=\dfrac{15}{45}\cdot\dfrac{1}{3}\cdot\dfrac{34}{7}=\dfrac{34}{63}\)
\(P=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}\)
\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)\)
\(=2\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=2\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=2.\dfrac{2}{5}=\dfrac{4}{5}\)
bn cộng hết lại vs nhau là ra kq!