Tính hợp lí: \(\dfrac{6}{21}-\dfrac{-12}{44}+\dfrac{10}{14}-\dfrac{1}{-4}-\dfrac{18}{33}\)
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=(2/7+5/7)+(3/11-6/11)+1/4
=5/4-3/11
=55/44-12/44=43/44
cho mk hỏi: 2/7, 5/7, 3/11, 6/11 ở đâu ra
mong nhận đc câu trả lời nhanh nhất ạ
=(2/7+5/7)+(3/11-6/11)+1/4
=5/4-3/11=55/44-12/44=43/44
6/21-(−12/44)+10/14−(1/(−4))−18/33
=2/7+12/44+5/7−((−1)/4)−6/11=2/7+12/44+5/7−((−1)/4)−6/11
=2/7+3/11+5/7+1/4−6/11=2/7+3/11+5/7+1/4−6/11
=(3/11−6/11)+(2/7+5/7)+1/4=(3/11−6/11)+(2/7+5/7)+1/4
=−3/11+7/7+1/4=−3/11+7/7+1/4
=43/44
a)P=\(\dfrac{-2}{7}+\dfrac{14}{29}+\dfrac{12}{33}+\dfrac{15}{29}+\dfrac{21}{33}+\dfrac{2}{7}\)
=\(\left(\dfrac{-2}{7}+\dfrac{2}{7}\right)+\left(\dfrac{14}{29}+\dfrac{15}{29}\right)+\left(\dfrac{12}{33}+\dfrac{21}{33}\right)\)
=0+1+1=2
b)\(\dfrac{2}{7}.\dfrac{5}{19}+\dfrac{2}{7}.\dfrac{14}{19}+\dfrac{21}{19}-\dfrac{2}{7}.\dfrac{1}{5}\)
=\(\dfrac{2}{7}.\left(\dfrac{5}{19}+\dfrac{14}{19}-\dfrac{1}{5}\right)+\dfrac{21}{19}\)
=\(\dfrac{2}{7}.\dfrac{4}{5}+\dfrac{21}{19}=\dfrac{887}{665}\)
\(1.\dfrac{-7}{18}+\dfrac{-5}{12}-\dfrac{-13}{18}\text{=}\left(\dfrac{-7}{18}-\dfrac{-13}{18}\right)+\dfrac{-5}{12}\text{=}\dfrac{1}{3}+\dfrac{-5}{12}\text{=}\dfrac{-1}{12}\)
\(2.\dfrac{-13}{17}+\dfrac{-13}{21}+\dfrac{-4}{17}\text{=}\left(\dfrac{-13}{17}+\dfrac{-4}{17}\right)+\dfrac{-13}{21}\text{=}-1+\dfrac{-13}{21}\text{=}\dfrac{-34}{21}\)
\(3.\dfrac{-13}{10}-\dfrac{-4}{13}+\dfrac{-11}{10}\text{=}\dfrac{-12}{5}-\dfrac{-4}{13}\text{=}\dfrac{-136}{65}\)
\(4.\dfrac{13}{17}\times\left(\dfrac{-4}{5}+\dfrac{-3}{4}\right)\text{=}\dfrac{13}{17}\times\dfrac{-31}{20}\text{=}\dfrac{-403}{340}\)
\(5.\left(\dfrac{-5}{12}\times\dfrac{-9}{20}\right)\times\dfrac{-7}{17}\text{=}\dfrac{3}{16}\times\dfrac{-7}{17}\text{=}\dfrac{-21}{272}\)
\(6.\dfrac{11}{23}\times\left(\dfrac{5}{9}+\dfrac{17}{9}-\dfrac{13}{9}\right)\text{=}\dfrac{11}{23}\times1\text{=}\dfrac{11}{23}\)
a) `6/18+ (-14)/21 = 1/3 - 2/3=-1/3`
b) `3/5 - (-1/2) = 3/5+1/2=11/10`
c) `2/7 : 3/4 = 2/7 . 4/3 = 8/21`
d) `(-28)/33 . (-3)/4 = 28/33 . 3/4=7/11`
a) \(\dfrac{6}{18}+\dfrac{-14}{21}=\dfrac{1}{3}-\dfrac{2}{3}=\dfrac{-1}{3}\)
b) \(\dfrac{3}{5}-\dfrac{-1}{2}=\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{6}{10}+\dfrac{5}{10}=\dfrac{11}{10}\)
c) \(\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}\cdot\dfrac{4}{3}=\dfrac{8}{21}\)
d) \(\dfrac{-28}{33}\cdot\dfrac{-3}{4}=\dfrac{28}{4}\cdot\dfrac{3}{33}=7\cdot\dfrac{1}{11}=\dfrac{7}{11}\)
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
= \(\dfrac{2}{7}\) - \(\dfrac{-3}{11}\)+\(\dfrac{5}{7}\)-\(\dfrac{-1}{4}\)- \(\dfrac{6}{11}\)
= (\(\dfrac{2}{7}\)+\(\dfrac{5}{7}\)) - ( \(\dfrac{-3}{11}\)+ \(\dfrac{6}{11}\)) - \(\dfrac{-1}{4}\)
= 1 - \(\dfrac{3}{11}\)-\(\dfrac{-1}{4}\)
= \(\dfrac{44}{44}\)-\(\dfrac{12}{44}\)-\(\dfrac{-11}{44}\)
= \(\dfrac{43}{44}\)