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Ta có :
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{30}{40}=\frac{3}{4}\)
\(\Rightarrow\)\(A>\frac{3}{4}\) ( điều phải chứng minh )
Vậy \(A>\frac{3}{4}\)
Chúc bạn học tốt ~
\(\left(2+\frac{5}{6}\right)\div1\frac{1}{5}+\frac{-7}{12}\)
\(=\left(\frac{12}{6}+\frac{5}{6}\right)\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\div\frac{6}{5}-\frac{7}{12}\)
\(=\frac{17}{6}\times\frac{5}{6}-\frac{7}{12}\)
\(=\frac{85}{12}-\frac{7}{12}\)
\(=\frac{78}{12}=\frac{13}{2}\)
\(\left(15-6\frac{13}{18}\right)\div11\frac{1}{7}-2\frac{1}{8}\div1\frac{11}{40}\)
\(=9\frac{13}{18}\div\frac{78}{7}-\frac{17}{8}\div\frac{51}{40}\)
\(=\frac{175}{18}\div\frac{78}{7}-\frac{17}{8}\times\frac{40}{51}\)
\(=\frac{175}{18}\times\frac{7}{78}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{5}{3}\)
\(=\frac{1225}{1404}-\frac{2340}{1404}\)
\(=\frac{-1115}{1404}\)
Bài 1:
a) Ta có: \(\frac{8}{40}+\frac{-4}{20}-\frac{3}{5}\)
\(=\frac{1}{5}+\frac{-1}{5}-\frac{3}{5}\)
\(=\frac{-3}{5}\)
b) Ta có: \(\frac{-7}{12}+\frac{-2}{12}-\frac{-3}{36}\)
\(=\frac{-7}{12}+\frac{-2}{12}-\frac{-1}{12}\)
\(=\frac{-9+1}{12}=\frac{-8}{12}=\frac{-2}{3}\)
c) Ta có: \(\left(\frac{1}{6}+\frac{-4}{13}\right)-\left(-\frac{17}{6}-\frac{30}{13}\right)\)
\(=\frac{1}{6}+\frac{-4}{13}+\frac{17}{6}+\frac{30}{13}\)
\(=3+2=5\)
d) Ta có: \(-\frac{-5}{4}+\frac{7}{4}-\frac{-11}{7}+\frac{2}{7}\)
\(=\frac{5}{4}+\frac{7}{4}+\frac{11}{7}+\frac{2}{7}\)
\(=3+\frac{13}{7}=\frac{21}{7}+\frac{13}{7}=\frac{34}{7}\)
e) Ta có: \(-\frac{1}{8}+\frac{-7}{9}+\frac{-7}{8}+\frac{6}{7}+\frac{2}{14}\)
\(=-1+1+\frac{-7}{9}\)
\(=-\frac{7}{9}\)
f) Ta có: \(\frac{-2}{9}-\frac{11}{-9}+\frac{5}{7}-\frac{-6}{-7}\)
\(=\frac{-2-\left(-11\right)}{9}+\frac{5-6}{7}\)
\(=1+\frac{-1}{7}=\frac{7}{7}+\frac{-1}{7}=\frac{6}{7}\)
b
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+..+\frac{1}{70}\)
Ta thấy:
\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( có 10 phân số \(\frac{1}{20}\)) = \(\frac{1}{20}\).10 = \(\frac{1}{2}\)
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}\)(có 10 phân số \(\frac{1}{30}\)) = \(\frac{1}{30}\).10 = \(\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( có 10 phân số \(\frac{1}{40}\)) = \(\frac{1}{40}\).10 = \(\frac{1}{4}\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}>\frac{1}{50}+\frac{1}{50}+...+\frac{1}{50}\)( có 10 phân số \(\frac{1}{50}\)) =\(\frac{1}{50}.10=\frac{1}{5}\)
\(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( có 10 phân số \(\frac{1}{60}\)) =\(\frac{1}{60}.10=\frac{1}{6}\)
\(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}>\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\)( có 10 phân số \(\frac{1}{70}\)) \(=\frac{1}{70}.10=\frac{1}{7}\)
=> A> \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}=\frac{223}{140}=\frac{699}{420}>\frac{560}{420}=\frac{4}{3}\)
=> A > \(\frac{4}{3}\)
\(a,\frac{7}{12}\cdot\frac{6}{11}+\frac{7}{12}\cdot\frac{5}{11}+2\frac{7}{12}\)
\(=\frac{7}{12}\cdot\left(\frac{6}{11}+\frac{5}{11}\right)+2\frac{7}{12}\)
\(=\frac{7}{12}+\frac{31}{12}\)
\(=\frac{38}{12}=\frac{19}{6}\)
\(b,\frac{-5}{9}\cdot\frac{-6}{13}+\frac{5}{-9}\cdot\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{-5}{9}\cdot\frac{-6}{13}+\frac{-5}{9}\cdot\frac{-5}{13}+\frac{-5}{9}\cdot1\)
\(=\frac{-5}{9}\cdot\left(\frac{-6}{13}+\frac{-5}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\left(\frac{-11}{13}+1\right)\)
\(=\frac{-5}{9}\cdot\frac{2}{13}\)
\(=\frac{-10}{117}\)
\(c,\)\(0,8\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}\cdot\frac{-15}{14}-\frac{4}{5}\cdot\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(\frac{-15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}\cdot\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
\(d,\)\(75\%\cdot\frac{6}{7}+5\%\cdot\frac{6}{7}+\frac{7}{10}\cdot1\frac{1}{7}\)
\(=\frac{3}{4}\cdot\frac{6}{7}+\frac{1}{20}\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\left(\frac{3}{4}+\frac{1}{20}\right)\cdot\frac{6}{7}+\frac{7}{10}\cdot\frac{8}{7}\)
\(=\frac{4}{5}\cdot\frac{6}{7}+\frac{4}{5}\cdot1\)
\(=\frac{4}{5}\cdot\left(\frac{6}{7}+1\right)\)
\(=\frac{4}{5}\cdot\frac{13}{7}\)
\(=\frac{52}{35}\)
a)7/12.6/11+7/12.5/11-2.7/12
=7/12(6/11+5/11-2)
=7/12(1-2)
=7/12.(-1)
=-7/12
a,\(\frac{-2}{3}+\frac{1}{5}=\frac{-2.5}{3.5}+\frac{3.1}{3.5}=-\frac{10}{15}+\frac{3}{15}=-\frac{7}{15}\)
b,\(3\frac{11}{13}-5\frac{11}{3}=3+\frac{11}{13}-(5+\frac{11}{13})=\left(3-5\right)+\left(\frac{11}{13}-\frac{11}{13}\right)=-2+0=-2\)
c,\(1\frac{2}{3}:(-\frac{5}{3})=\frac{5}{3}:\left(-\frac{5}{3}\right)=\frac{5.3}{3.\left(-5\right)}=\frac{15}{-15}=-1\)
d,\(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)=1+\left(-1\right)=0\)
Tìm x:
a,x+12=8
x=8-12
x=-4
b,\(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{5}{10}\)
\(\frac{2}{3}x=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{2}{3}\)
\(x=\frac{6}{10}\)
\(x=\frac{3}{5}\)
a) \(\frac{-2}{3}+\frac{1}{5}=\frac{-10}{15}+\frac{3}{15}=\frac{-10+3}{15}=\frac{-7}{15}\)
b) \(3\frac{11}{13}-5\frac{11}{13}=\frac{50}{13}-\frac{76}{13}=\frac{50-76}{13}=\frac{-26}{13}=-2\)
c) \(1\frac{2}{3}:\frac{-5}{3}=\frac{5}{3}:\frac{-5}{3}=\frac{5}{3}\times\frac{3}{-5}=\frac{15}{-15}=-1\)
d) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)
\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)
\(=\left(\frac{31-14}{17}\right)+\left(\frac{-5-8}{13}\right)\)
\(=\frac{17}{17}+\frac{-13}{13}\)
\(=1+\left(-1\right)\)
\(=0\)
TÌM x
a) \(x+12=8\)
\(\Leftrightarrow x=8-12\)
\(\Leftrightarrow x=-4\)
b) \(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{1-5}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{-4}{10}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}:\frac{2}{3}=\frac{-2}{5}\times\frac{3}{2}\)
\(\Leftrightarrow x=\frac{-6}{10}=\frac{-3}{5}\)