NT
5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
24 tháng 7 2023
a 25 phần 41
b 9
c 1 phần 12
d 12
e 14 phần 15
f 24 phần 7
28 tháng 2 2022
a: -3/4x12=-36/4=-9
b: \(=\dfrac{7}{15}\cdot\dfrac{10}{21}=\dfrac{7}{21}\cdot\dfrac{10}{15}=\dfrac{1}{3}\cdot\dfrac{2}{3}=\dfrac{2}{9}\)
c: \(=\dfrac{7}{18}\cdot\dfrac{12}{21}=\dfrac{12}{18}\cdot\dfrac{7}{21}=\dfrac{2}{3}\cdot\dfrac{1}{3}=\dfrac{2}{9}\)
d: \(=\dfrac{11}{15}\cdot\dfrac{5}{22}=\dfrac{11}{22}\cdot\dfrac{5}{15}=\dfrac{1}{2}\cdot\dfrac{1}{3}=\dfrac{1}{6}\)
e: \(=-\dfrac{8}{15}\cdot\dfrac{5}{4}=\dfrac{-40}{60}=\dfrac{-2}{3}\)
f: \(=-15\cdot\dfrac{3}{5}=-9\)
10 tháng 5 2020
a, \(\frac{3}{5}+\frac{-4}{15}=\frac{9}{15}-\frac{4}{15}=\frac{5}{15}=\frac{1}{3}\)
b, \(\frac{-1}{3}+\frac{2}{5}+\frac{2}{15}=\frac{-5}{15}+\frac{6}{15}+\frac{2}{15}=\frac{3}{15}=\frac{1}{5}\)
c, \(\frac{-3}{5}+\frac{7}{21}+\frac{-4}{5}+\frac{7}{5}=\frac{-3}{5}+\frac{1}{3}+\frac{-4}{5}+\frac{7}{5}=\left(\frac{-3}{5}+\frac{-4}{5}+\frac{7}{5}\right)+\frac{1}{3}=\frac{1}{3}\)
d, \(\frac{2}{7}+\frac{1}{9}+\frac{3}{7}+\frac{5}{9}+\frac{-5}{6}=\left(\frac{2}{7}+\frac{3}{7}\right)+\left(\frac{1}{9}+\frac{5}{9}\right)+\frac{-5}{6}=\frac{5}{7}+\frac{6}{9}+\frac{-5}{6}=\frac{90}{126}+\frac{84}{126}+\frac{-105}{126}=\frac{69}{126}=\frac{23}{42}\)
e, \(\frac{-5}{7}+\frac{3}{4}+\frac{-1}{5}+\frac{-2}{7}+\frac{1}{4}=\left(\frac{-5}{7}+\frac{-2}{7}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{-1}{5}=\left(-1\right)+1+\frac{-1}{5}=\frac{-1}{5}\)
f, \(\frac{-3}{31}+\frac{-6}{17}+\frac{1}{25}+\frac{-28}{31}+\frac{-1}{17}+\frac{-1}{5}=\left(\frac{-3}{31}+\frac{-28}{31}\right)+\left(\frac{-6}{17}+\frac{-1}{17}\right)+\left(\frac{1}{25}+\frac{-1}{5}\right)=\left(-1\right)+\frac{-7}{17}+\frac{-4}{25}=\frac{-425}{425}+\frac{-175}{425}+\frac{-68}{425}=\frac{-668}{425}\)
Chúc bn học tốt
KN
10 tháng 2 2019
\(\frac{7}{8}\div\left(\frac{14}{3}+\frac{14}{4}\right)+\frac{10}{14}\)
\(=\frac{7}{8}\div\left(\frac{56}{12}+\frac{52}{12}\right)+\frac{10}{14}\)
\(=\frac{7}{8}\div\frac{108}{12}+\frac{10}{14}\)
\(=\frac{7}{8}\div9+\frac{5}{7}\)
\(=\frac{7}{8}\times\frac{1}{9}+\frac{5}{7}\)
\(=\frac{7}{72}+\frac{5}{7}\)
\(=\frac{49}{504}+\frac{360}{504}\)
\(=\frac{409}{504}\)
\(\frac{30}{42}\div\left(\frac{5}{7}-\frac{5}{21}\right)+\frac{6}{8}\)
\(=\frac{6}{7}\div\left(\frac{15}{21}-\frac{5}{21}\right)+\frac{3}{4}\)
\(=\frac{6}{7}\div\frac{10}{21}+\frac{3}{4}\)
\(=\frac{6}{7}\times\frac{21}{10}+\frac{3}{4}\)
\(=\frac{126}{70}+\frac{3}{4}\)
\(=\frac{504}{280}+\frac{210}{280}\)
\(=\frac{714}{280}\)
KN
10 tháng 2 2019
\(\left(6-\frac{14}{15}\right).\frac{50}{38}-\frac{8}{18}\)
\(=\left(\frac{90}{15}-\frac{14}{15}\right).\frac{25}{19}-\frac{4}{9}\)
\(=\frac{76}{15}.\frac{25}{19}-\frac{4}{9}\)
\(=\frac{76.25}{15.19}-\frac{4}{9}\)
\(=\frac{19.2.2.5.5}{3.5.19}-\frac{4}{9}\)
\(=\frac{20}{3}-\frac{4}{9}\)
\(=\frac{60}{9}-\frac{4}{9}\)
\(=\frac{56}{9}\)
\(\frac{4}{6}+\frac{5}{21}.\left(\frac{4}{5}-\frac{2}{6}\right)\)
\(=\frac{2}{3}+\frac{5}{21}.\left(\frac{24}{30}-\frac{10}{30}\right)\)
\(=\frac{2}{3}+\frac{5}{21}.\frac{14}{30}\)
\(=\frac{2}{3}+\frac{70}{630}\)
\(=\frac{420}{630}+\frac{70}{630}\)
\(=\frac{490}{630}\)
\(=\frac{7}{9}\)
CK
3
TD
2 tháng 2 2022
\(\dfrac{-7}{8}:\dfrac{21}{16}-\dfrac{5}{3}.\left(\dfrac{1}{5}-\dfrac{7}{10}\right)\)
\(=\dfrac{-7}{8}.\dfrac{16}{21}-\dfrac{5}{3}.\left(\dfrac{-1}{2}\right)\)
\(=\dfrac{-2}{3}+\dfrac{5}{6}=\dfrac{1}{6}\)
\(\frac{-3}{5}+\frac{7}{21}+\frac{-4}{5}+\frac{7}{5}\)
=\(\left(\frac{-3}{5}+\frac{-4}{5}+\frac{7}{5}\right)+\frac{7}{21}\)
=\(\left(\frac{-7}{5}+\frac{7}{5}\right)+\frac{7}{21}\)
=0+\(\frac{7}{21}\)
=\(\frac{7}{21}\)
=\(\frac{1}{3}\)
bo tay @gmail.com