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\(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+....+\frac{1}{20}\)
\(=\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}\right)\)
\(>\frac{1}{15}\cdot5+\frac{1}{20}\cdot5\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}>\frac{6}{12}=\frac{1}{2}\)
\(\Rightarrow S>\frac{1}{2}\)
Bài làm
Ta có:
\(\frac{1}{11}>\frac{1}{20}\), \(\frac{1}{12}>\frac{1}{20}\), \(\frac{1}{13}>\frac{1}{20}\), \(\frac{1}{14}>\frac{1}{20}\), \(\frac{1}{15}>\frac{1}{20}\), \(\frac{1}{16}>\frac{1}{20}\), \(\frac{1}{17}>\frac{1}{20}\), \(\frac{1}{18}>\frac{1}{20}\),\(\frac{1}{19}>\frac{1}{20}\)
=> \(S=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}>\frac{1}{20}\)
hay \(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+\frac{1}{20}\)
=> \(S=\frac{1}{20}.10=\frac{10}{20}=\frac{1}{2}\)
Do đó: \(S=\frac{1}{2}\)
# Chúc bạn học tốt #
\(x.\frac{3}{7}=\frac{2}{3}\)
\(x=\frac{2}{3}:\frac{3}{7}\)
\(x=\frac{14}{9}\)
\(24:\frac{-6}{11}=24.\frac{-11}{6}\)
\(=\frac{-264}{6}=-44\)
Bài 1:
a) b) c) sẽ có bạn giải cho em thôi vì nó dễ tính tay cũng đc
d) \(\frac{4}{2.5}+\frac{4}{5.8}+...+\frac{4}{23.26}\)
\(=\frac{4}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{23.26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{23}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{2}-\frac{1}{26}\right)\)
\(=\frac{4}{3}.\frac{6}{13}\)
\(=\frac{8}{13}\)
Bài 2:
a) b) c)
d)\(|\frac{5}{8}x+\frac{6}{7}|-\frac{4}{7}=\frac{10}{7}\)
\(\Leftrightarrow|\frac{5}{8}x+\frac{6}{7}|=2\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{6}{7}=2\\\frac{5}{8}x+\frac{6}{7}=-2\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{8}{7}\\\frac{5}{8}x=\frac{-20}{7}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{64}{35}\\x=\frac{-32}{7}\end{cases}}}\)
Vậy \(x\in\left\{\frac{64}{35};\frac{-32}{7}\right\}\)
Bài 1 :
a) \(\left(\frac{2}{5}-\frac{5}{8}\right):\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-9}{40}:\frac{11}{30}+\frac{1}{8}\)
\(=\frac{-27}{44}+\frac{1}{8}\)
\(=\frac{-43}{88}\)
M=17/5 x (-31)/125 x 1/2 x 10/17 x 1/23
M=(17/5 x 1/2)x(-31/125 x 1/23)x 10/7
M=17/10 x (-31/125 x 1/8)x 10/7
M=(17/10 x 10/17)x(-31/1000)
M=170/170 x (-31/1000)
M=1 x (-31/1000)
M= -31/1000
a)\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)vaB=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
+)Ta có:\(A=\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{2}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\left(\frac{7}{32}+\frac{128}{32}\right)\)
\(\Leftrightarrow A=\frac{31}{23}-\frac{135}{32}\)
\(\Leftrightarrow A=\frac{992}{736}-\frac{3105}{736}\)
\(\Leftrightarrow A=\frac{-2113}{736}\left(1\right)\)
+)Ta lại có:\(B=\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(\frac{12}{67}-\frac{79}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)
\(\Leftrightarrow B=\frac{1}{3}+\frac{-67}{67}+\frac{41}{41}\)
\(\Leftrightarrow B=\frac{1}{3}+\left(-1\right)+1\)
\(\Leftrightarrow B=\frac{1}{3}\left(2\right)\)
+)Từ (1) và (2)
\(\Leftrightarrow A< 0< B\Leftrightarrow A< B\)
Vậy A<B
b)\(\frac{200420042004}{200520052005}va\frac{2004}{2005}\)
+)Ta có \(\frac{200420042004}{200520052005}=\frac{2004.100010001}{2005.100010001}=\frac{2004}{2005}\)
\(\Leftrightarrow\frac{200420042004}{200520052005}=\frac{2004}{2005}\)
c)\(C=\frac{2020^{2006}+1}{2020^{2007}+1}vaD=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(C=\frac{2020^{2006}+1}{2020^{2007}+1}< 1\)
\(\Leftrightarrow C< \frac{2020^{2006}+1+2019}{2020^{2007}+1+2019}=\frac{2020^{2006}+2020}{2020^{2007}+2020}=\frac{2020.\left(2020^{2005}+1\right)}{2020.\left(2020^{2006}+1\right)}=\frac{2020^{2005}+1}{2020^{2006}+1}\)
\(\Leftrightarrow C< D\)
Chúc bạn học tốt
Lê Minh Phương tham khảo bài mình nhé
\(a,\frac{9}{-7}< x>\frac{7}{2}\)
\(\Leftrightarrow\frac{-9}{7}< x>\frac{7}{2}\)
\(\Leftrightarrow\frac{-18}{14}< x>\frac{49}{14}\)
\(\Leftrightarrow-18< x>49\)
\(\Leftrightarrow x\in\left\{-17;-16;-15;...;50\right\}\)
Còn bài kia tương tự
\(a,\frac{9}{-7}< x< \frac{7}{2}\)
\(\Rightarrow\frac{9.2}{-7.2}< x< \frac{7.7}{2.7}\)
\(\Rightarrow\frac{-18}{14}< x< \frac{49}{14}\)
\(\text{vì}x\in Z\Rightarrow x=-\frac{14}{14};\frac{0}{14};\frac{14}{14};\frac{28}{14};\frac{42}{14}\)
\(\text{hay }x=\left\{-1;0;1;2;3\right\}\)
\(5\frac{1}{7}=\frac{36}{7}\)
\(6\frac{3}{4}=\frac{27}{4}\)
\(1\frac{12}{13}=\frac{25}{13}\)
(\(\frac{12}{19}.\frac{19}{12}\))(\(\frac{-13}{17}.\frac{17}{13}\))\(\frac{7}{15}\)
= 1 . (-1) . \(\frac{7}{15}\)
= \(-\frac{7}{15}\)
Ta có: \( \left(\frac{12}{19}\times\frac{19}{12}\right)\times\left(\frac{-13}{17}\times\frac{17}{13}\right).\frac{19}{12}\)
=1.(-1).\(\frac{19}{12}\)
=(-1).\(\frac{19}{12}\)
=\(-\frac{19}{12}\)