Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
giúp mình bài này với
bạn nam đọc 1 quyển sách trong 3 ngày . ngày 1 đọc 1 phần 3 số trang . ngày thứ 2 đọc 2 phần 5 số trang còn lại .ngày 3 đọc 36 trang còn lại. hỏi quyển sách có bao nhiêu trang
\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)
= \(1\dfrac{11}{25}+\dfrac{5}{7}\)
= \(2\dfrac{27}{175}\)
b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)
=\(-2+\left(\dfrac{ }{ }\right)\)
a) Ta có: \(-19-\left|+13\right|+\left|-10\right|-\left(-5\right)-11\)
\(=-19-13+10+5-11\)
\(=-28\)
b) Ta có: \(\left(-10\right)-\left(11\right)+\left|-5\right|-\left|-7\right|-8\)
\(=-10-11+5-7-8\)
\(=-31\)
c) Ta có: \(-\left(-8\right)-\left(+7\right)-\left|-11\right|+\left|+12\right|\)
\(=8-7-11+12\)
\(=2\)
d) Ta có: \(-14-\left|-18\right|-\left(-20\right)-\left|-25\right|\)
\(=-14-18+20-25\)
\(=-37\)
đ) Ta có: \(23-\left|-15\right|-\left(-17\right)+\left|-13\right|\)
\(=23-15+17+13\)
\(=38\)
Bài 1:
a; \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{7}{21}\) + (- \(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= \(\dfrac{5}{18}\) + \(\dfrac{8}{19}\) - \(\dfrac{1}{3}\) -\(\dfrac{10}{36}\) + \(\dfrac{11}{19}\) + \(\dfrac{1}{3}\) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{10}{36}\)) + (\(\dfrac{8}{19}\) + \(\dfrac{11}{19}\)) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\)) - \(\dfrac{5}{8}\)
= (\(\dfrac{5}{18}\) - \(\dfrac{5}{18}\)) + \(\dfrac{19}{19}\) - 0 - \(\dfrac{5}{8}\)
= 0 + 1 - \(\dfrac{5}{8}\)
= \(\dfrac{3}{8}\)
b; \(\dfrac{1}{13}\) + (\(\dfrac{-5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\)) - (\(\dfrac{12}{17}\) - \(\dfrac{5}{18}\) + \(\dfrac{7}{5}\))
= \(\dfrac{1}{13}\) - \(\dfrac{5}{18}\) - \(\dfrac{1}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{12}{17}\) + \(\dfrac{5}{18}\) - \(\dfrac{7}{5}\)
= (\(\dfrac{1}{13}\) - \(\dfrac{1}{13}\)) + (\(\dfrac{12}{17}\) - \(\dfrac{12}{17}\)) + (-\(\dfrac{5}{18}\) + \(\dfrac{5}{18}\)) - \(\dfrac{7}{5}\)
= 0 + 0 + 0 - \(\dfrac{7}{5}\)
= - \(\dfrac{7}{5}\)
Bài 1 c;
\(\dfrac{15}{14}\) - (\(\dfrac{17}{23}\) - \(\dfrac{80}{87}\) + \(\dfrac{5}{4}\)) + (\(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\))
= \(\dfrac{15}{14}\) - \(\dfrac{17}{23}\) + \(\dfrac{80}{87}\) - \(\dfrac{5}{4}\) + \(\dfrac{17}{23}\) - \(\dfrac{15}{14}\) + \(\dfrac{1}{4}\)
= (\(\dfrac{15}{14}-\dfrac{15}{14}\)) + (\(-\dfrac{17}{23}+\dfrac{17}{23}\)) - (\(\dfrac{5}{4}\) - \(\dfrac{1}{4}\)) + \(\dfrac{80}{87}\)
= 0 + 0 - 1 + \(\dfrac{80}{87}\)
= - \(\dfrac{7}{87}\)
\(\frac{7}{25}+\frac{-18}{25}+\frac{4}{23}+\frac{5}{7}+\frac{19}{23}\)
\(=\left(\frac{7}{25}-\frac{11}{25}\right)+\left(\frac{4}{23}+\frac{19}{23}\right)+\frac{5}{7}\)
\(=\frac{-4}{25}+1+\frac{5}{7}\)
\(=\frac{-28}{175}+\frac{175}{175}+\frac{125}{175}\)
\(=\frac{272}{175}\)
a) 10-x-5=-5-7-11
=> 5 - x = -23
=> x = 28
b) |x| -3=0
=> |x| = 3
=> x = 3 hoặc x -3
c) ( 7-|x| ) .(2x-4)=0
=> 7 - |x| = 0 hoặc 2x - 4 = 0
=> |x| = 7 hoặc 2x = 4
=> x = 7 hoặc x = - 7 hoặc x = 2
c)2+3x=-15-19
=> 2 + 3x = -34
=> 3x = 36
=> x = 12
\(a,\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=\left(-1\right)+1+\dfrac{5}{7}\)
\(=0+\dfrac{5}{7}=\dfrac{5}{7}\)
b, \(\dfrac{7}{19}.\dfrac{8}{11}+\dfrac{7}{19}.\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}.1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}=1\)
a. \(\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=-1+1+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\)
b. \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\cdot1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}\)
\(=1\)