Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{5}-\dfrac{49}{50}=\dfrac{-48}{50}=-\dfrac{24}{25}\)
501−(1−21+21−31+...+491−501)
=15−4950=−4850=−2425=51−5049=50−48=−2524
Ta có: \(P=\frac{1}{99}-\frac{1}{99\cdot98}-\frac{1}{98\cdot97}-\frac{1}{97\cdot96}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(=\frac{1}{99}-\left(\frac{1}{98}-\frac{1}{99}\right)-\left(\frac{1}{97}-\frac{1}{98}\right)-\left(\frac{1}{96}-\frac{1}{97}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)
\(=\frac{1}{99}-\frac{1}{98}+\frac{1}{99}-\frac{1}{97}+\frac{1}{98}-\frac{1}{96}+\frac{1}{97}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(=\frac{2}{99}-1=\frac{2}{99}-\frac{99}{99}\)
\(=\frac{-97}{99}\)
a) x=8,25-19/6 b)-5/8-x=3/20+1/6 c)x+1/4=-40/48+6/48 d)x=-42/35.25/-14 K
x=33/4-19/6 -5/8-x=9/60+10/60 x+1/4=-17/24 x=3/7.5/1
x=99/12-38/12 x=-5/8-19/60 x=-17/24-1/4 x=15/7
x=61/12 x=-75/120-38/120 x=-17/24-6/24
x=-113/120 x=-23/24
a ) \(8,25-x=3\frac{1}{6}\)
\(\frac{33}{4}-x=\frac{19}{6}\)
\(x=\frac{33}{4}-\frac{19}{6}\)
\(x=\frac{99}{12}-\frac{38}{12}\)
\(x=\frac{61}{12}\)
Vậy \(x=\frac{61}{12}\)
b ) \(-\frac{5}{8}-x=\frac{3}{20}-\left(-\frac{1}{6}\right)\)
\(-\frac{5}{8}-x=\frac{3}{20}+\frac{1}{6}\)
\(-\frac{5}{8}-x=\frac{9}{60}+\frac{10}{60}\)
\(-\frac{5}{8}-x=\frac{19}{60}\)
\(x=-\frac{5}{8}-\frac{19}{60}\)
\(x=-\frac{75}{120}-\frac{38}{120}\)
\(x=-\frac{113}{120}\)
Vậy \(x=-\frac{113}{120}\)
c ) \(x-\left(-\frac{1}{4}\right)=-\frac{5}{6}+\frac{1}{8}\)
\(x+\frac{1}{4}=-\frac{20}{24}+\frac{3}{24}\)
\(x+\frac{1}{4}=-\frac{17}{24}\)
\(x=-\frac{17}{24}-\frac{1}{4}\)
\(x=-\frac{17}{24}-\frac{6}{24}\)
\(x=-\frac{23}{24}\)
Vậy \(x=-\frac{23}{24}\)
d ) \(-\frac{14}{25}x=-\frac{42}{25}\)
\(x=-\frac{42}{25}:-\frac{14}{25}\)
\(x=-\frac{42}{25}.-\frac{25}{14}\)
\(x=3\)
Vậy \(x=3\)
Chúc bạn học tốt !!!

x^1 x^2 à bạn
\(2x-4x-6x=-8.25\)
\(-12x=-8,25\)
\(\Rightarrow x=\frac{8.25}{12}=0,6875\)