\(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{3}-125\frac{3}{5}\)

\(=\left|-\frac{419}{15}\right|+\left(-\frac{419}{15}\right)\)

\(=\frac{419}{15}+\left(-\frac{419}{15}\right)=0\)

học tốt ~~

a, \(139\frac{5}{7}:\frac{2}{3}−138\frac{2}{7}:\sqrt{\frac{4}{9}} \)

= \(139\frac{5}{7}:\frac{2}{3}−138\frac{2}{7}:\frac{2}{3}\)

= \((139\frac{5}{7}−138\frac{2}{7}):\frac{2}{3}\)

= \(1\frac{3}{7}:\frac{2}{3}\)

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

b, \((\frac{-5}{11}:\frac{13}{18}-\frac{5}{11}:\frac{13}{5})+\frac{-1}{33} \)

= \((\frac{5}{11}.\frac{-18}{13}-\frac{5}{11}.\frac{5}{13})+\frac{-1}{33}\)

= \([\frac{5}{11}.(\frac{-18}{13}-\frac{5}{13})]+\frac{-1}{33}\)

= \((\frac{5}{11}.\frac{-23}{13})+\frac{-1}{33}\)

= \(\frac{-155}{143}+\frac{-1}{33}\)

= \(\frac{-358}{429} \)

c, \(∣97\frac{2}{3}-125\frac{3}{5}∣+97\frac{2}{3}-125\frac{3}{5} \)

= \(∣\frac{-419}{15}∣+97\frac{2}{3}-125\frac{3}{5}\)

= \(\frac{419}{15}+97\frac{2}{3}-125\frac{3}{5}\)

= \(0\)

Tick cho mình nha!!!

Chúc bạn học tốt.

\(\frac{\frac{125}{8}+\frac{125}{97}+\frac{125}{576}+\frac{250}{991}}{\frac{25}{8}+\frac{25}{97}+\frac{25}{576}+\frac{50}{991}}\)=\(\frac{250.\left(\frac{1}{8}+\frac{1}{97}+\frac{1}{576}+\frac{1}{991}\right)}{50.\left(\frac{1}{8}+\frac{1}{97}+\frac{1}{576}+\frac{1}{991}\right)}\)=\(\frac{250}{50}\)=5

Nhớ tick

Lời giải:

** Sửa đề: Chỗ $\frac{1}{1}$ ở mẫu chuyển thành $\frac{1}{2}$

$\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+....+\frac{1}{97}.3+\frac{1}{99}.1$

$=50+(\frac{97}{3}+1)+(\frac{95}{5}+1)+....+(\frac{3}{97}+1)+(\frac{1}{99}+1)$

$=50+\frac{100}{3}+\frac{100}{5}+...+\frac{100}{97}+\frac{100}{99}$
$=100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})$

\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})}=\frac{1}{100}\)