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22 tháng 11 2021

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} đây là biểu thức gì\)

1 tháng 8 2018

a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)

  =\(\frac{3}{8}+\frac{2}{13}\)

  =\(\frac{55}{104}.\)

b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)

  =\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)

  =\(\frac{2}{7}+\frac{2}{7}\)

  =\(\frac{4}{7}\)

c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)

  =\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)

  =\(-\frac{16}{55}--\frac{1}{50}\)

  =\(-\frac{149}{550}.\)

d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)

  =\(-\frac{417}{1564}\)

e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)

  =\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)

  =\(\frac{380}{391}.\)

g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)

  =\(-\frac{5}{28}+-\frac{132}{85}\)

  = \(-1.731512605.\)

k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................

5 tháng 8 2018

THANK  Ngô Bùi Hoa  làm cho mình bài 2 với 

13 tháng 3 2019

hỏi chị google ấy

13 tháng 3 2019

A= \(\frac{1}{31}.\left[\frac{5}{31}\left(9-\frac{1}{2}\right)-\frac{17}{2}\left(4+\frac{1}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

\(\frac{1}{31}.\left(\frac{5}{31}.\frac{17}{2}-\frac{17}{2}.\frac{21}{5}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{5}{31}-\frac{21}{5}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left[\frac{17}{2}.\left(\frac{-626}{155}\right)\right]+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{1}{31}.\left(\frac{-5321}{155}\right)+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{-5321}{4805}+\frac{1}{2}+\frac{1}{6}+...+\frac{1}{930}\)

=\(\frac{-5321}{4805}+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{30.31}\)

=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{30}-\frac{1}{31}\)

=\(\frac{-5321}{4805}+\frac{1}{1}-\frac{1}{31}\)

=\(\frac{-5321}{4805}+\frac{30}{31}\)

=\(\frac{-671}{4805}\)

21 tháng 7 2018

a)   \(\frac{31}{23}-\left(\frac{7}{32}+\frac{8}{23}\right)=\frac{31}{23}-\frac{7}{32}-\frac{8}{23}=1-\frac{7}{32}=\frac{25}{32}\)

b)   \(\left(\frac{1}{3}+\frac{12}{67}+\frac{13}{41}\right)-\left(\frac{79}{67}-\frac{28}{41}\right)\)

\(=\frac{1}{3}+\frac{12}{67}+\frac{13}{41}-\frac{79}{67}+\frac{28}{41}\)

\(=\frac{1}{3}-\left(\frac{79}{67}-\frac{12}{67}\right)+\left(\frac{13}{41}+\frac{28}{41}\right)\)

\(=\frac{1}{3}-1+1=\frac{1}{3}\)

d)   \(\frac{1}{7}.\frac{1}{3}+\frac{1}{7}.\frac{-1}{3}+\frac{17}{19}=\frac{1}{7}.\left(\frac{1}{3}-\frac{1}{3}\right)+\frac{17}{19}=\frac{17}{19}\)

e)  \(\frac{3}{5}.\frac{7}{9}+\frac{7}{5}.\frac{2}{9}=\frac{7}{5}.\left(\frac{3}{9}+\frac{2}{9}\right)=\frac{7}{5}.\frac{5}{9}=\frac{7}{9}\)

28 tháng 8 2020

b)

\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5.1\frac{31}{45}\right):\left(21.\frac{1}{2}\right)\)

\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7< x< \left(\frac{16}{5}:\frac{16}{5}+\frac{9}{2}.\frac{76}{45}\right):\frac{21}{2}\)

\(\Rightarrow2-7< x< \left(1+\frac{38}{5}\right):\frac{21}{2}\)

\(\Rightarrow-5< x< \frac{43}{5}:\frac{21}{2}\)

\(\Rightarrow-5< x< \frac{86}{105}\)

\(x\in Z\left(gt\right)\)

\(\Rightarrow x\in\left\{-4;-3;-2;-1;0\right\}.\)

Vậy \(x\in\left\{-4;-3;-2;-1;0\right\}.\)