\(\dfrac{5}{9}\) . \(\dfrac{10}{11}\) + \(\dfrac{5}{9}\) . \(\dfrac{14}{11}\) - \(\dfrac{5}{9}\) . \(\dfrac{15}{11}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
8: \(=\dfrac{-5}{9}-\dfrac{4}{9}+\dfrac{8}{15}+\dfrac{7}{15}-\dfrac{2}{11}=\dfrac{-2}{11}\)
9: =2/7-2/5+5/7=1-2/5=3/5
10: \(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)-\dfrac{12}{19}=\dfrac{-5}{19}\)
11: \(=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{-5}{7}\)
Mik làm Bài 2 nhé ~
Bài 2 :
a) \(x-\dfrac{1}{2}=-\dfrac{1}{10}\)
\(x=-\dfrac{1}{10}+\dfrac{1}{2}\)
\(x=\dfrac{2}{5}\)
b) \(\dfrac{2}{3}x-\dfrac{7}{6}=\dfrac{5}{2}\)
\(\dfrac{2}{3}x=\dfrac{5}{2}+\dfrac{7}{6}\)
\(\dfrac{2}{3}x=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}:\dfrac{2}{3}\)
\(x=\dfrac{11}{3}.\dfrac{3}{2}\)
\(x=\dfrac{11}{2}\)
c) \(2,5-\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{3}{4}\)
\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=2,5-\dfrac{3}{4}\)
\(\left(\dfrac{1}{8}x+\dfrac{1}{2}\right)=\dfrac{5}{2}-\dfrac{3}{4}\)
\(\dfrac{1}{8}x+\dfrac{1}{2}=\dfrac{7}{4}\)
\(\dfrac{1}{8}x=\dfrac{7}{4}-\dfrac{1}{2}\)
\(\dfrac{1}{8}x=\dfrac{5}{4}\)
\(x=10\)
Bài 1:
a) \(\dfrac{-4}{11}.\dfrac{7}{9}+\dfrac{-4}{11}.\dfrac{2}{9}-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}.\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}.1-\dfrac{7}{11}\)
\(=\dfrac{-4}{11}-\dfrac{7}{11}\)
\(=-1\)
b) \(\dfrac{3}{5}:\dfrac{-7}{10}+0,5-\left(\dfrac{-9}{14}\right)\)
\(=\dfrac{-6}{7}+\dfrac{1}{2}+\dfrac{9}{14}\)
\(=\dfrac{2}{7}\)
c) \(\dfrac{3}{5}-\dfrac{8}{5}:\left(5,25+75\%\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:\left(\dfrac{21}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:6\)
\(=\dfrac{3}{5}-\dfrac{4}{15}\)
\(=\dfrac{1}{3}\)
−5/9. 15 11/13−(−5/9).11 11/13+5/9
= −5/9 . (15 11/13 − 11 11/13) + 5/9
= −5/9 . 4 + 5/9
= −20/9 + 5/9
= −5/3
Like cho mình nha
\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)
a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
A=0+0+0+...+0+\(\dfrac{13}{15}\)
A=\(\dfrac{13}{15}\)
b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)
\(a.\)
\(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{5\cdot3-5\cdot2}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
\(b.\)
\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2-5-9}{11}=-\dfrac{12}{11}\)
\(c.\)
\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{5}{12}+\dfrac{1}{15}=\dfrac{6-5\cdot5+4}{60}=-\dfrac{15}{60}=-\dfrac{1}{4}\)
a) \(\dfrac{5}{16}-\dfrac{5}{24}=\dfrac{15}{48}-\dfrac{10}{48}=\dfrac{15-10}{48}=\dfrac{5}{48}\)
b)\(\dfrac{2}{11}+\left(\dfrac{-5}{11}-\dfrac{9}{11}\right)=\dfrac{2}{11}-\dfrac{5}{11}-\dfrac{9}{11}=\dfrac{2-5-9}{11}=\dfrac{-12}{11}\)
c)\(\dfrac{1}{10}-\left(\dfrac{5}{12}-\dfrac{1}{15}\right)=\dfrac{1}{10}-\dfrac{7}{20}=\dfrac{2}{20}-\dfrac{7}{20}=\dfrac{-5}{20}=\dfrac{-1}{4}\)
\(A=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)-\left(\dfrac{3}{4}+\dfrac{2}{9}+\dfrac{1}{36}\right)+\dfrac{1}{64}\)
\(=\dfrac{5+9+1}{15}-\dfrac{27+8+1}{36}+\dfrac{1}{64}\)
=1/64
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
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!!! ~
=\(\dfrac{5}{11}\) nhé
cho mik xin 1like nhé:>>