\(|97\frac{2}{3}-125\frac{3}{5}|+97\frac{2}{5}-125\frac{1}{3}\)
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\(\left|97\frac{2}{3}-125\frac{3}{5}\right|+97\frac{2}{3}-125\frac{3}{5}\)
\(=\left|-\frac{419}{15}\right|+\left(-\frac{419}{15}\right)\)
\(=\frac{419}{15}+\left(-\frac{419}{15}\right)=0\)
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
Lời giải:
** Sửa đề: Chỗ $\frac{1}{1}$ ở mẫu chuyển thành $\frac{1}{2}$
$\frac{1}{1}.99+\frac{1}{3}.97+\frac{1}{5}.95+....+\frac{1}{97}.3+\frac{1}{99}.1$
$=50+(\frac{97}{3}+1)+(\frac{95}{5}+1)+....+(\frac{3}{97}+1)+(\frac{1}{99}+1)$
$=50+\frac{100}{3}+\frac{100}{5}+...+\frac{100}{97}+\frac{100}{99}$
$=100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})$
\(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}}{100(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99})}=\frac{1}{100}\)
Ta có: \(\left|97\dfrac{2}{3}-125\dfrac{3}{5}\right|+97\dfrac{2}{5}-125\dfrac{1}{3}\)
\(=\left|97+\dfrac{2}{3}-125-\dfrac{3}{5}\right|+97+\dfrac{2}{5}-125-\dfrac{1}{3}\)
\(=\left|-28+\dfrac{1}{15}\right|-28+\dfrac{1}{15}\)
\(=\left|\dfrac{1}{15}-28\right|-28+\dfrac{1}{15}\)
\(=28-\dfrac{1}{15}-28+\dfrac{1}{15}\)
\(=0\)
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được