Cho biểu thức E=\(\dfrac{\dfrac{4}{11}+\dfrac{4}{121}-\dfrac{4}{12321}}{\dfrac{9}{11}+\dfrac{9}{121}-\dfrac{9}{12321}}\)
Chứng tỏ rằng E là một số hữu tỉ
Chứng tỏ rằng giá trị các biểu thức là 1 số hữu tỉ A =6/71/2 B = 4/15 nhân -25/24 C =0,3 nhân 12,8+0,3 nhân 7,2 D= 1/10 nhân 11+1/11+12+…+1/99 nhân 100 E =4/11+4/121-4/12321/9/11+9/121-9/12321
\(\dfrac{3}{15}.\dfrac{5}{9}:\dfrac{-18}{17}.\dfrac{14}{17}\)
\(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
giúp mình với
a: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{-14}{18\cdot9}=\dfrac{-14}{162}=\dfrac{-7}{81}\)
b: \(=\dfrac{12}{4}+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
\(\dfrac{3}{15}.\dfrac{5}{9}:\dfrac{-18}{17}.\dfrac{14}{17}\)
\(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
giúp mình với
a: \(=\dfrac{15}{135}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{1}{9}\cdot\dfrac{-7}{9}=\dfrac{-7}{81}\)
b: \(=3+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
Bài 1: Thực hiện phép tính:
a, \(\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right)\)
b, \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\)
c, \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
Tính một cách hợp lý:
a\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{100}-1\right)\)) \(x:\dfrac{99}{100}:\dfrac{98}{99}:...:\dfrac{2}{3}:\dfrac{1}{2}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}\)
c) \(\dfrac{\dfrac{1}{9}-\dfrac{5}{6}-4}{\dfrac{7}{12}-\dfrac{1}{36}-10}\)
d) \(\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)...\left(\dfrac{1}{99}+1\right)\)
e)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
\(\dfrac{1}{5}\)+\(\dfrac{1}{13}\)+\(\dfrac{1}{25}\)+...+\(\dfrac{1}{10^2}\)+\(\dfrac{1}{11^2}\)< \(\dfrac{9}{20}\)
Chứng tỏ rằng biểu thức trên bé hơn 9/20
Bài 15:
a)\(\dfrac{-2}{5}\)+\(\dfrac{4}{5}\) . x =\(\dfrac{3}{5}\)
b)\(\dfrac{-3}{7}\) - \(\dfrac{4}{7}\):x = -2
Bài 16
a) x - \(\dfrac{10}{3}\) = \(\dfrac{7}{15}\) . \(\dfrac{3}{5}\)
b) x + \(\dfrac{3}{22}\)= \(\dfrac{27}{121}\) . \(\dfrac{11}{9}\)
c) \(\dfrac{8}{23}\) . \(\dfrac{48}{24}\) - x = \(\dfrac{1}{3}\)
d) 1 - x = \(\dfrac{49}{65}\).\(\dfrac{5}{7}\)
Bài 17: tìm x
a) \(\dfrac{62}{7}\) . x = \(\dfrac{29}{9}\): \(\dfrac{3}{56}\)
b) \(\dfrac{1}{5}\) : x=\(\dfrac{1}{5}\)+\(\dfrac{1}{7}\)
bài 18:
a)\(\dfrac{2}{5}\)+\(\dfrac{3}{4}\): x =\(\dfrac{-1}{2}\)
b)\(\dfrac{5}{7}\) - \(\dfrac{2}{3}\) . x = \(\dfrac{4}{5}\)
c) \(\dfrac{1}{2}\)x + \(\dfrac{3}{5}\)x = \(\dfrac{-2}{3}\)
d) \(\dfrac{4}{7}\).x-x = \(\dfrac{-9}{14}\)
bài 19: tính
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+ \(\dfrac{1}{2018.2019}\)
bài 20:tìm x
\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{x.\left(x+1\right)}\)=\(\dfrac{2008}{2009}\)
bài 21: tìm x
\(\dfrac{x+1}{99}\)+\(\dfrac{x+2}{98}\)\(\dfrac{x+3}{97}\)\(\dfrac{x+4}{96}\)=-4
bài 22 : so sánh các phân số sau:
a) \(\dfrac{-1}{5}\)+\(\dfrac{4}{-5}\)và 1
b) \(\dfrac{3}{5}\) và \(\dfrac{2}{3}\)+\(\dfrac{-1}{5}\)
c)\(\dfrac{3}{2}\)+\(\dfrac{-4}{3}\) và \(\dfrac{1}{10}\)+\(\dfrac{-4}{5}\)
d) \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{5}\)+\(\dfrac{1}{6}\) và 2
\(\dfrac{3}{4}.\dfrac{7}{9}.\dfrac{1}{9}.\dfrac{7}{4}\)
\(\dfrac{6}{7}.\dfrac{8}{13}-\dfrac{6}{9}.\dfrac{9}{7}+\dfrac{5}{13}.\dfrac{6}{7}\)
2.11.\(\dfrac{3}{4}.\dfrac{9}{121}\)
\(\dfrac{3}{4}\cdot\dfrac{7}{9}\cdot\dfrac{1}{9}\cdot\dfrac{7}{4}\)
\(=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot9\cdot9\cdot4}=\dfrac{3\cdot7\cdot1\cdot7}{4\cdot3\cdot3\cdot9\cdot4}\)
\(=\dfrac{7\cdot1\cdot7}{4\cdot3\cdot9\cdot4}=\dfrac{49}{432}\)
\(\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{9}\cdot\dfrac{9}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\cdot\dfrac{8}{13}-\dfrac{6}{7}+\dfrac{5}{13}\cdot\dfrac{6}{7}\)
\(=\dfrac{6}{7}\left(\dfrac{8}{13}-1+\dfrac{5}{13}\right)\)
\(=\dfrac{6}{7}\cdot0\)
\(=0\)
\(2\cdot11\cdot\dfrac{3}{4}\cdot\dfrac{9}{121}\)
\(=\dfrac{2\cdot11\cdot3\cdot9}{4\cdot121}=\dfrac{2\cdot11\cdot3\cdot9}{2\cdot2\cdot11\cdot11}\)
\(=\dfrac{3\cdot9}{2\cdot11}=\dfrac{27}{22}\)
a. \(\dfrac{3.7.1.7}{4.3.3.9.4}=\dfrac{49}{432}\)
b. \(\dfrac{6}{7}.\left(\dfrac{8}{13}+\dfrac{5}{13}-1\right)=\dfrac{6}{7}.0=0\)
\(\dfrac{2.11.3.9}{2.2.11.11}=\dfrac{27}{22}\)
\(\)a) \(\dfrac{3}{5}+\dfrac{11}{20}\) b) \(\dfrac{5}{8}-\dfrac{4}{9}\) c) \(\dfrac{9}{16}\) x \(\dfrac{4}{3}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}\) e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}\)
a) \(\dfrac{3}{5}+\dfrac{11}{20}=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b) \(\dfrac{5}{8}-\dfrac{4}{9}=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c) \(\dfrac{9}{16}\times\dfrac{4}{3}=\dfrac{3}{4}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{4}{5}\times\dfrac{5}{2}=\dfrac{3}{5}+2=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
a)\(=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b)\(=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c)\(=\dfrac{9\times4}{16\times3}=\dfrac{3}{4}\)
d)\(=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e)\(=\dfrac{3}{5}+\dfrac{4}{2}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)