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\(\dfrac{7}{-25}+\dfrac{18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
=\(\left(\dfrac{-7}{25}+\dfrac{18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
= \(\dfrac{11}{25} +1+\dfrac{5}{7}\)
= \(1\dfrac{11}{25}+\dfrac{5}{7}\)
= \(2\dfrac{27}{175}\)
b) \(-2+\dfrac{15}{19}+\dfrac{-15}{17}+\dfrac{15}{23}+\dfrac{4}{19}\)
=\(-2+\left(\dfrac{ }{ }\right)\)
\(a,\dfrac{7}{-25}+\dfrac{-18}{25}+\dfrac{4}{23}+\dfrac{5}{7}+\dfrac{19}{23}\)
\(=\left(\dfrac{-7}{25}+\dfrac{-18}{25}\right)+\left(\dfrac{4}{23}+\dfrac{19}{23}\right)+\dfrac{5}{7}\)
\(=\left(-1\right)+1+\dfrac{5}{7}\)
\(=0+\dfrac{5}{7}=\dfrac{5}{7}\)
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. \(\dfrac{7}{19}\cdot\dfrac{8}{11}+\dfrac{7}{19}\cdot\dfrac{3}{11}+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\left(\dfrac{8}{11}+\dfrac{3}{11}\right)+\dfrac{12}{19}\)
\(=\dfrac{7}{19}\cdot1+\dfrac{12}{19}\)
\(=\dfrac{7}{19}+\dfrac{12}{19}\)
\(=1\)
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
2)
\(2+\dfrac{5}{7}+\left(\dfrac{\dfrac{3}{19}+\dfrac{3}{23}-\dfrac{3}{28}}{\dfrac{5}{19}+\dfrac{5}{23}-\dfrac{5}{28}}\right)\cdot x=\dfrac{20}{7}\\ \left[\dfrac{3\cdot\left(\dfrac{1}{19}+\dfrac{1}{23}-\dfrac{1}{28}\right)}{5\cdot\left(\dfrac{1}{19}+\dfrac{1}{23}-\dfrac{1}{28}\right)}\right]\cdot x=\dfrac{20}{7}-\dfrac{5}{7}-2\\ \dfrac{3}{5}x=\dfrac{15}{7}-2\\ \dfrac{3}{5}x=\dfrac{1}{7}\\ x=\dfrac{5}{21}\)
\(\dfrac{-5}{19}+\dfrac{7}{23}+\dfrac{-19}{19}\)
\(=\left(\dfrac{-5}{19}+\dfrac{-19}{19}\right)+\dfrac{7}{23}\)
\(=\dfrac{-24}{19}+\dfrac{7}{23}\)
\(=\dfrac{-552}{437}+\dfrac{133}{437}\)
\(=\dfrac{-419}{437}\)
Lời giải:
$\frac{-5}{19}+\frac{7}{23}+\frac{-19}{19}=\frac{7}{23}-\frac{5}{19}-1$
$=\frac{7.19-5.23}{23.19}-1$
$=\frac{18}{437}-1=\frac{-419}{437}$