Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)
b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)
d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)
e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)
\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)
f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)
g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)
h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)
i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)
1) âm năm phần 12
2) âm mười bảy phần 9
3) -1
Đây là đáp án còn làm bài từ làm nhé
b: =12+5/14-3-5/7-5-5/14
=4-5/7
=28/7-5/7=23/7
c: =(-2/5-11/10)+(7/11-7/11)
=-4/10-11/10=-15/10=-3/2
\(a,\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{5}{13}\cdot\dfrac{2}{9}\)
\(=\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{2}{13}\cdot\dfrac{5}{9}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}+\dfrac{8}{13}-\dfrac{2}{13}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{14}{13}\)
\(=\dfrac{70}{117}\)
\(d,\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}+\dfrac{-2}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}\right)+\dfrac{-2}{5}\)
\(=0+\dfrac{-2}{5}\)
\(=\dfrac{-2}{5}\)
a) Ta có: \(\left|5\cdot0.6+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=\left|3+\dfrac{2}{3}\right|-\dfrac{1}{3}\)
\(=3+\dfrac{2}{3}-\dfrac{1}{3}\)
\(=3+\dfrac{1}{3}=\dfrac{10}{3}\)
b) Ta có: \(\left(0.25-1\dfrac{1}{4}\right):5-\dfrac{1}{5}\cdot\left(-3\right)^2\)
\(=\left(\dfrac{1}{4}-\dfrac{5}{4}\right)\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{-4}{4}\cdot\dfrac{1}{5}-\dfrac{1}{5}\cdot9\)
\(=\dfrac{1}{5}\cdot\left(-1-9\right)\)
\(=-10\cdot\dfrac{1}{5}=-2\)
c) Ta có: \(\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}:\dfrac{5}{7}\)
\(=\dfrac{14}{17}\cdot\dfrac{7}{5}-\dfrac{-3}{17}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot\left(\dfrac{14}{17}+\dfrac{3}{17}\right)\)
\(=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)
d) Ta có: \(\dfrac{7}{16}+\dfrac{-9}{25}+\dfrac{9}{16}+\dfrac{-16}{25}\)
\(=\left(\dfrac{7}{16}+\dfrac{9}{16}\right)-\left(\dfrac{9}{25}+\dfrac{16}{25}\right)\)
\(=\dfrac{16}{16}-\dfrac{25}{25}\)
\(=1-1=0\)
e) Ta có: \(\dfrac{5}{6}+2\sqrt{\dfrac{4}{9}}\)
\(=\dfrac{5}{6}+2\cdot\dfrac{2}{3}\)
\(=\dfrac{5}{6}+\dfrac{4}{3}\)
\(=\dfrac{5}{6}+\dfrac{8}{6}=\dfrac{13}{6}\)
\(\dfrac{5}{3}\cdot\dfrac{7}{25}+\dfrac{5}{3}\cdot\dfrac{21}{25}-\dfrac{5}{3}\cdot\dfrac{7}{25}\)
\(=\dfrac{5}{3}\cdot\left(\dfrac{7.}{25}+\dfrac{21}{25}-\dfrac{7}{25}\right)\)
\(=\dfrac{5}{3}\cdot\dfrac{21}{25}=\dfrac{7}{5}\)
b) \(250\%+19\dfrac{3}{11}\cdot\dfrac{7}{26}-6\dfrac{3}{11}\cdot\dfrac{7}{26}\)
\(=\dfrac{5}{2}+\dfrac{212}{11}\cdot\dfrac{7}{26}-\dfrac{69}{11}\cdot\dfrac{7}{26}\)
\(=\dfrac{7}{26}\cdot\left(\dfrac{212}{11}-\dfrac{69}{11}\right)+\dfrac{5}{2}\)
\(=\dfrac{7}{26}\cdot13+\dfrac{5}{2}\)
\(=\dfrac{7}{2}+\dfrac{5}{2}\)
\(=\dfrac{12}{2}=6\)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
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!!! ~
\(=2-\left(\dfrac{5}{3}-\dfrac{7}{6}+\dfrac{9}{10}-...-\dfrac{19}{45}\right)\)
\(=2-2\left(\dfrac{5}{6}-\dfrac{7}{12}+\dfrac{9}{20}-...-\dfrac{19}{90}\right)\)
\(=2-2\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{5}-...-\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=2-2\cdot\dfrac{4}{10}=2-\dfrac{8}{10}=2-\dfrac{4}{5}=\dfrac{6}{5}\)
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) \(\dfrac{3}{5}-\dfrac{8}{5}:\left(5,25+75\%\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:\left(\dfrac{21}{4}+\dfrac{3}{4}\right)\)
\(=\dfrac{3}{5}-\dfrac{8}{5}:6\)
\(=\dfrac{3}{5}-\dfrac{4}{15}\)
\(=\dfrac{1}{3}\)
a)\(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}\\ =\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}\\ =\dfrac{14-15+1}{35}\\ =\dfrac{0}{35}=0\)
b)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\\ =\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}\\ =-1+1+\dfrac{2}{7}\\ =0+\dfrac{2}{7}=\dfrac{2}{7}\)