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\(C=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{13}+...+\frac{1}{2011}.\frac{1}{2015}\)
\(C=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.13}+...+\frac{1}{2011.2015}\)
\(4C=4\left(\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.13}+...+\frac{1}{2011.2015}\right)\)
\(4C=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.13}+...+\frac{4}{2011.2015}\)
\(4C=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{2011}-\frac{1}{2015}\)
\(4C=\frac{1}{3}-\frac{1}{2015}=\frac{2012}{6045}\)
\(C=\frac{2012}{6045}:4=\frac{503}{6045}\)
\(C=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{15}+....+\dfrac{1}{2011}-\dfrac{1}{2015}\right)\)
\(C=\dfrac{1}{4}\left(\dfrac{1}{3}-\dfrac{1}{2015}\right)=\dfrac{1}{4}\left(\dfrac{2012}{3.2015}\right)=\dfrac{503}{3.2015}\)
b1 -10/14
b2 -4/5
b3
a 2/9-7/8.x=1/3
7/8.x=2/9-1/3=-1/9
x=-1/9:-7/8=8/63
b 23/7.x-1/8=11/4
23/7.x=11/4+1/8=23/8
x=23/8:23/7=7/8
b4
Quyển truyện cs số trang:
36:(1−1/4−9/20)=120(trang)
Đây này má Ran mori
a) \(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
\(=5+\dfrac{1}{7}-3-\dfrac{3}{11}-2-\dfrac{1}{7}-1-\dfrac{8}{11}\)
\(=\left(5-3-2-1\right)+\left(\dfrac{1}{7}-\dfrac{3}{11}-\dfrac{1}{7}-\dfrac{8}{11}\right)\)
\(=-1+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\)
\(=-1+0-1=-2\)
a)\(\left(5\dfrac{1}{7}-3\dfrac{3}{11}\right)-2\dfrac{1}{7}-1\dfrac{8}{11}\)
= \(\left(5+\dfrac{1}{7}-3+\dfrac{3}{11}\right)-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(5-\dfrac{1}{7}+3-\dfrac{3}{11}-2+\dfrac{1}{7}-1+\dfrac{8}{11}\)
= \(\left(5-3-2-1\right)+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{8}{11}-\dfrac{3}{11}\)
= \(-1+2+\dfrac{5}{11}\)
= \(1+\dfrac{5}{11}=\dfrac{1}{1}+\dfrac{5}{11}=\dfrac{11}{11}+\dfrac{5}{11}=\dfrac{16}{11}\)
Vậy :câu a) = \(\dfrac{16}{11}\)
a: \(=\left(15-6-\dfrac{13}{18}\right):\dfrac{298}{27}-\dfrac{17}{8}:\dfrac{51}{40}\)
\(=\dfrac{149}{18}\cdot\dfrac{27}{298}-\dfrac{5}{3}=\dfrac{3}{2}-\dfrac{5}{3}=\dfrac{9-10}{6}=\dfrac{-1}{6}\)
b: \(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\dfrac{-22}{15}:\dfrac{11}{2}\)
\(=\dfrac{3}{4}-\dfrac{4}{15}=\dfrac{29}{60}\)
c: \(=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+5+\dfrac{7}{9}=\dfrac{-7}{9}+\dfrac{7}{9}+5=5\)
d: \(=\dfrac{1}{2}\cdot\dfrac{4}{3}\cdot10\cdot\dfrac{1}{5}\cdot\dfrac{3}{4}=1\)
e: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}+\dfrac{-23}{4}=\dfrac{204}{25}\)
a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)
b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)
\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)
c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)
d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)
