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S
30 tháng 3 2017
a, \(\dfrac{7}{18}x-\dfrac{2}{3}=\dfrac{5}{18}\)
\(\dfrac{7}{18}x=\dfrac{5}{18}+\dfrac{2}{3}\)
\(\dfrac{7}{18}x=\dfrac{17}{18}\)
\(x=\dfrac{17}{18}\div\dfrac{7}{18}\)
\(x=\dfrac{17}{7}\)
C
30 tháng 3 2017
a) \(\dfrac{7}{18}\).x-\(\dfrac{2}{3}\)=\(\dfrac{5}{18}\)
\(\dfrac{7}{18}\).x =\(\dfrac{5}{18}\)+\(\dfrac{2}{3}\)
\(\dfrac{7}{18}\).x = \(\dfrac{17}{18}\)
x = \(\dfrac{17}{18}\) :\(\dfrac{7}{18}\)
x =\(\dfrac{17}{7}\)
b)\(\dfrac{4}{9}\) - \(\dfrac{7}{8}\).x =\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{4}{9}\)-\(\dfrac{-2}{3}\)
\(\dfrac{7}{8}\).x =\(\dfrac{10}{9}\)
x =\(\dfrac{10}{9}\) : \(\dfrac{7}{8}\)
x =\(\dfrac{80}{63}\)
c)\(\dfrac{1}{6}\)+\(\dfrac{-5}{7}\): \(\dfrac{-7}{18}\)
\(\dfrac{1}{6}\)+\(\dfrac{90}{49}\)
\(\dfrac{589}{294}\)
NM
2
31 tháng 1 2018
a) 1-2-3+4+5-6-7+8+......+17-18-19+20=(1-2-3+4)+(5-6-7+8)+....+(17-18-19+20)
=0+0+......+0
=0
câu b hơi khó mình cần thời gian ,mai mình đưa kết quả cho
9 tháng 8 2021
a)\(\frac{-5}{18}+\frac{32}{45}-\frac{9}{10}\)
\(=\frac{-25}{90}+\frac{64}{90}-\frac{81}{90}\)
\(=-\frac{42}{90}=-\frac{7}{15}\)
b)\(\frac{3}{4}-\left(-\frac{5}{8}\right)+\frac{9}{4}-\frac{3}{16}\)
\(=\frac{3}{4}+\frac{5}{8}+\frac{9}{4}-\frac{3}{16}\)
\(=\frac{12}{16}+\frac{10}{16}+\frac{36}{16}-\frac{3}{16}\)
\(=\frac{58}{16}=\frac{29}{8}\)
c) \(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}=\frac{11}{125}\)
d) \(\frac{-2}{5}-\frac{3}{10}-\left(-\frac{1}{2}\right)\)
\(=-\frac{2}{5}-\frac{3}{10}+\frac{1}{2}\)
\(=-\frac{4}{10}-\frac{3}{10}+\frac{5}{10}\)
\(=-\frac{2}{10}=-\frac{1}{5}\)
TL
5
9 tháng 8 2017
A = 4 . 2 . 25 . 5 . 175
A = 22 . 2 . 52 . 5 . 52 . 7
A = 22+1 . 52+1+2 . 7
A = 23 . 55 .7
A = 175000
H
9 tháng 8 2018
\(B=4^2-10^4:\left(50\cdot273-50\cdot73\right)\)
\(B=4^2-10^4:\left[50\cdot\left(273-73\right)\right]\)
\(B=4^2-10^4:\left(50\cdot200\right)\)
\(B=4^2-10^4:10^4=4^2-1=15\)
\(C=3\times53\times6+2\times9\times87-18\times40\)
\(C=18\times53+18\times87-18\times40\)
\(C=18\times\left(53+87-40\right)\)
\(C=18\times100=1800\)
\(C=\dfrac{4}{3.5}+\dfrac{4}{5.7}+\dfrac{4}{7.9}+...+\dfrac{4}{97.99}\)
\(=2\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}\right)\)
\(=2\left(\dfrac{1}{3}-\dfrac{1}{99}\right)\)
\(=2\cdot\dfrac{32}{99}=\dfrac{64}{99}\)
\(D=\dfrac{18}{2.5}+\dfrac{18}{5.8}+...+\dfrac{18}{203.206}\)
\(=6\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{203.206}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{203}-\dfrac{1}{206}\right)\)
\(=6\left(\dfrac{1}{2}-\dfrac{1}{206}\right)\)
\(=6\cdot\dfrac{51}{103}=\dfrac{306}{103}\)
\(\Rightarrow\dfrac{C}{D}=\dfrac{\dfrac{64}{99}}{\dfrac{306}{103}}=\dfrac{3296}{15147}\)
ai trả lời giúp mik với