\(\frac{2}{4}+\frac{5}{4}=\frac{7}{4}\)

= \(\frac{7}{4}\).

~~~~~~ MK NHA ~~~~~~~~

3(x+2)=-4(x-5)

3x+6=-4x+20

3x+4x=20-6

7x=14

x=14:7=2

ta có: 3/x - 5 = -4/x + 2

     => 3(x + 2) = -4(x + 5)

     => 3x + 6 = (-4)x + -20

     => 3x + 6 = (-4)x - 20

     => (-4)x - 3x = 20 + 6

     => (-1)x = 26

     => x = -26

=\(\frac{49}{12}\)

nha bn

ủng hộ và k cho mk nha

Quy Đồng Ra

=\(\frac{17}{5}\)

\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{16}\right)+...+1+\frac{1}{1024}\)

\(=\left(1+1+1+...+1\right)+\left(1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{1024}\right)\)

\(=9+\frac{1}{512}=\frac{4609}{512}\)

\(A=\frac{2-\frac{2}{19}+\frac{2}{43}-\frac{2}{1995}}{3-\frac{3}{19}+\frac{3}{43}-\frac{3}{1995}}:\frac{4-\frac{4}{29}+\frac{4}{41}-\frac{4}{2941}}{5-\frac{5}{29}+\frac{5}{41}-\frac{5}{2941}}\)

\(\Rightarrow A=\frac{2\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1995}\right)}{3\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1995}\right)}:\frac{4\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}{5\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}\)

\(\Rightarrow A=\frac{2}{3}:\frac{4}{5}\)

\(\Rightarrow A=\frac{2}{3}.\frac{5}{4}\)

\(\Rightarrow A=\frac{10}{12}=\frac{5}{6}\)

thanks bạn

\(\frac37+\frac{5}{13}+\frac{4}{13}\)

=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)

=\(\frac37+\frac{9}{13}\)

=\(\frac{39}{91}+\frac{63}{91}\)

=\(\frac{102}{91}\)

\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)

=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)

=\(\frac{5}{24}\times\frac65+\frac12\)

=\(\frac14+\frac12\)

=\(\frac14+\frac24\)

=\(\frac34\)

Gọi \(A=\frac{5}{4.6}+\frac{5}{6.8}+\frac{5}{8.10}+...+\frac{5}{298.300}\)

\(\Rightarrow A=5\left[\frac{1}{2}\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{298}-\frac{1}{300}\right)\right]\)

\(\Rightarrow A=\frac{5}{2}.\left(\frac{1}{4}-\frac{1}{300}\right)\)

\(\Rightarrow A=\frac{37}{60}\)

Bạn ơi cho mình hỏi \(\frac{1}{2}\)ở đâu có vậy

Ta có: \(1\frac{4}{5}+2\frac{5}{7}+3\frac{4}{5}+4\frac{5}{7}\)

\(=\left(1\frac{4}{5}+3\frac{4}{5}\right)+\left(2\frac{5}{7}+4\frac{5}{7}\right)\)

\(=\left(\frac{9}{5}+\frac{19}{5}\right)+\left(\frac{19}{7}+\frac{33}{7}\right)\)

\(=\frac{28}{5}+\frac{52}{7}=13\frac{1}{35}\)

= ( \(1\frac{4}{5}\)+ \(3\frac{4}{5}\)) + (  \(2\frac{5}{7}\)+  \(4\frac{5}{7}\))         

=        \(4\frac{4}{5}\)           +       \(6\frac{5}{7}\)

=           \(\frac{24}{5}\)         +        \(\frac{47}{7}\)

=  ...... ( tính nốt nhé )