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Ta có: \(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2013.2015}\right)\)
\(=2\left(2+1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2013}-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{2}{2015}\)
Vậy D < 6.
\(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+...+\frac{\left(2!\right)^2}{2015^2}\)
=>\(D=\frac{\left(1.2\right)^2}{1^2}+\frac{\left(1.2\right)^2}{3^2}+\frac{\left(1.2\right)^2}{5^2}+\frac{\left(1.2\right)^2}{7^2}+...+\frac{\left(1.2\right)^2}{2015^2}\)
=>\(D=\frac{2^2}{1^2}+\frac{2^2}{3^2}+\frac{2^2}{5^2}+\frac{2^2}{7^2}+...+\frac{2^2}{2015^2}\)
=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)\)
Ta có: \(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}< 2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\)
=>\(D=2\left(\frac{2}{1^2}+\frac{2}{3^2}+\frac{2}{5^2}+\frac{2}{7^2}+...+\frac{2}{2015^2}\right)< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)
Mà \(2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)\)\(=2\left(2+\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=2\left(2+1-\frac{1}{2015}\right)=2\left(3-\frac{1}{2015}\right)=6-\frac{6}{2016}< 6\)
=>\(D< 2\left(2+\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2013.2015}\right)< 6\)
=>D<6
=\(\frac{3}{4}-\frac{5}{2}-\frac{5}{3}+\left(\frac{-1}{8}\right)=\frac{18}{24}-\frac{60}{24}-\frac{40}{24}-\frac{3}{24}=\frac{-85}{24}\)
\(\frac{3}{4}-\left(\frac{5}{2}+\frac{5}{3}\right)+\left(\frac{-1}{2}\right)^3\)
\(=\frac{3}{4}-\frac{25}{6}+\frac{-1}{8}\)
\(=\frac{18-100-3}{24}\)
\(=\frac{-85}{24}\)
Tk mk nha!
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
x-0,27=(73/77+73/165+73/185):[25.12/133)
x-0,27=(73/77+73/165+73/185):[25/1.12/133)
x-0,27=(73/77+73/165+73/185):300/133
x-0,27=1387/777:300/133
x-0,27=0,791381381
x=0,791381381-0,27
x=0,521381381
a) (x-3)+(x-2)+(x-1)+....+10+11=11
(x-3)+(x-2)+(x-1)+....+10 =0
gọi số hạng của tổng vế trái là n
(x-3+10).\(\frac{n}{2}\)=0
(x+7).n:2=0
(x+7) =0
\(\Rightarrow\)x+7=0 (do n\(\ne\)0)
x=0-7
x=-7
b) \(\frac{2}{3}\left[\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right]<=x<=4\frac{1}{3}.\left[\frac{1}{2}-\frac{1}{6}\right]\)
\(\frac{2}{3}.\frac{11}{12}<=x<=\frac{13}{3}.\frac{1}{3}\)
\(\frac{11}{18}<=x<=\frac{13}{9}\)
do x\(\in\)z nên x=1
vậy x=1
\(\frac{1}{3}x+\frac{2}{5}\left(x-1\right)=0\)
\(\frac{1}{3}x+\frac{2}{5}x-\frac{2}{5}=0\)
\(x\left(\frac{1}{3}+\frac{2}{5}\right)=\frac{2}{5}\)
\(x.\frac{11}{15}=\frac{2}{5}\)
x=6/11
=\(\frac{4}{9}\)nhé
\(\left(-\frac{2}{3}\right)^2=\frac{\left(-2\right)^2}{3^2}=\frac{4}{9}\)