\(\frac{1.3.5+2.6.10+4.12.20+7.21.35}{1.5.7+2.10.14+4.20.28+7.35.49}\)

\(=\frac{1.3.5\left(1+2+4+7\right)}{1.5.7\left(1+2+7+7\right)}=\frac{1.3.5}{1.5.7}=\frac{15}{35}=\frac{3}{7}\)

\(\frac{1\cdot3\cdot5+2\cdot6\cdot10+4\cdot12\cdot20+7\cdot21\cdot35}{1\cdot5\cdot7+2\cdot10\cdot14+4\cdot20\cdot28+7\cdot35\cdot49}\)

\(=\)\(\frac{1\cdot3\cdot5\cdot\left(1+2+4+7\right)}{1\cdot5\cdot7\cdot\left(1+2+7+7\right)}\)

\(=\frac{1\cdot3\cdot5}{1\cdot5\cdot7}\)\(=\frac{15}{35}=\frac{3}{7}\)

bn vội quá viết nhầm lun kìa 

hj hj chúc bn làm bài tốt nha

\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)

\(=\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)

\(=\frac{59}{15}-\left[\frac{7}{3}\left(\frac{9}{2}-2\right)\right]:\frac{7}{4}\)

\(=\frac{59}{15}-\frac{35}{6}:\frac{7}{4}\)

\(=\frac{59}{15}-\frac{10}{3}\)

\(=\frac{9}{15}=\frac{3}{5}\)

\(\cdot62,87+35,14+4,13+8,35+4,86+5,65\)

\(=\left(62,87+4,13\right)+\left(35,14+4,86\right)+\left(8,35+5,65\right)\)

\(=67+40+14\)

\(=121\)

a) \(\frac{51}{3}-\frac{22}{3}=\frac{51-22}{3}=\frac{29}{3}\)

b) \(\frac{5}{12}+\frac{5}{6}-\frac{3}{4}=\frac{5}{12}+\frac{10}{12}-\frac{9}{12}=\frac{5+10-9}{12}=\frac{6}{12}=\frac{1}{2}\)

c) \(1-\left(\frac{1}{5}+\frac{1}{2}\right)=\frac{10}{10}-\frac{2}{10}-\frac{5}{10}=\frac{10-5-2}{10}=\frac{3}{10}\)

d) \(\frac{111}{4}-\left(\frac{25}{7}+\frac{51}{4}\right)=\frac{777}{28}-\frac{60}{28}-\frac{357}{28}=\frac{360}{28}=\frac{90}{7}\)

e) \(\left(\frac{85}{11}+\frac{35}{7}\right)-\frac{35}{11}=\left(\frac{85}{11}-\frac{35}{11}\right)+\frac{35}{7}=\frac{50}{11}-\frac{35}{7}=\frac{350}{77}-\frac{385}{77}=-\frac{35}{77}\)

cho mình hỏi 5 1/3 là hỗn số hả bạn hay nhân 5 vs 1/3

cho mình xin lỗi những số đó là hỗn số nhé

\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}=\frac{15}{93}\)

\(2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\right)=2.\frac{15}{93}\)

\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}=\frac{30}{93}\)

\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 93 - 3

=> 2x = 90

=> x = 90 : 2

=> x = 45

Vậy x = 45

sai rồi

Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right).\left(2x+3\right)}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\)

\(2A=\frac{1}{3}-\frac{1}{2x+3}\)

\(A=\frac{2x}{3\left(2x+3\right)}:2\)

\(\Rightarrow\frac{2x}{3\left(2x+3\right)}:2=\frac{15}{93}\)

\(\frac{2x}{3\left(2x+3\right)}=\frac{15}{93}.2=\frac{30}{93}=\frac{10}{31}\)

\(\frac{2x}{6x+9}=\frac{10}{31}\)

\(\Rightarrow2x.31=10.\left(6x+9\right)\)

\(\Rightarrow62x=60x+90\)

\(62x-60x=90\)

\(2x=90\)

\(x=45\)

Vậy x = 45

Ủng hộ mk nha !!! ^_^

Ta có : \(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+.....+\frac{3}{9.11}\)

\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{9.11}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

Ta có :

\(\frac{3}{1.3}+\frac{3}{3.5}+........+\frac{3}{9.11}\)

\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+.....+\frac{3}{9.11}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}\left(1-\frac{1}{11}\right)\)

\(=\frac{3}{2}.\frac{10}{11}=\frac{30}{22}\)

đề bài là j bạn?? 

so sánh à bạn 

là tính tổng á

\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)

\(10\dfrac{1}{5}-5\dfrac{1}{2}\times\dfrac{60}{11}+3\div15\%\\ =\dfrac{51}{5}-\dfrac{11}{2}\times\dfrac{60}{11}+3\div\dfrac{15}{100}\\ =\dfrac{51}{5}-30+20\\ =10,2-30+20\\ =0,2\)

\(\dfrac{5}{7}\times\dfrac{5}{11}+\dfrac{5}{7}\times\dfrac{2}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times-\dfrac{7}{11}=-\dfrac{5}{11}\)

\(-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+1\dfrac{5}{7}\\ =-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times1+\dfrac{12}{7}\\ =1\)