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

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

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

\(\text{=}10+\dfrac{8}{5}+\dfrac{10}{7}\text{=}131\dfrac{1}{35}\)

16𝑥2=3𝑦+143

-0,086418888

A\(\frac{7}{8}+\frac{7}{12}+\frac{9}{8}+\frac{5}{12}\)

\(=\left(\frac{7}{8}+\frac{9}{8}\right)+\left(\frac{7}{12}+\frac{5}{12}\right)\)

\(=1+2\)

\(=3\)

B\(\frac{2}{5}\times\frac{4}{7}-\frac{3}{7}\times\frac{2}{5}\)

\(=\left(\frac{4}{7}-\frac{3}{7}\right)\times\frac{2}{5}\)

\(=\frac{1}{7}\times\frac{2}{5}\)

\(=\frac{2}{35}\)

ok

\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{100\cdot103}=\frac{1}{3}\left[\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right]\)

\(\frac{1}{3}\left[1-\frac{1}{103}\right]=\frac{1}{3}\cdot\frac{102}{103}=\frac{34}{103}\)

viết lại đề đi bạn

\(\frac{1}{1\cdot4}\)+\(\frac{1}{4\cdot7}\)+\(\frac{1}{7\cdot10}\)+...+\(\frac{1}{100\cdot103}\)

a) 45/28

b)62/525

c)49/90

`4 / 7 xx [-5] / 6 xx 7 / 4 xx 12 / 5`

`= [ 4 xx (-5) xx 7 xx 12 ] / [ 7 xx 6 xx 4 xx 5 ]`

`= [ 4 xx (-5) xx 7 xx 6 xx 2 ] / [ 7 xx 6 xx 4 xx 5 ]`

`= -2`

4/7 × -5/7 × 7/4 ×12/5 

=> -10/21 × 42/10

=> -420/210

=> -42/21

Đáp số: -42/21

A\(\frac{7}{8}+\frac{7}{12}+\frac{9}{8}+\frac{5}{12}\)

\(=\left(\frac{7}{12}+\frac{5}{12}\right)+\left(\frac{7}{8}+\frac{9}{8}\right)\)

\(=1+2\)

\(=3\)