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a) \(\dfrac{11}{21}+\dfrac{-4}{7}=\dfrac{11}{21}+\dfrac{-12}{21}=\dfrac{-1}{21}\)
b) \(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}=\dfrac{1}{3}+\dfrac{14}{25}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\left(\dfrac{14}{25}+\dfrac{11}{25}\right)+\dfrac{2}{7}=-1+1+\dfrac{2}{7}=\dfrac{2}{7}\)
c) \(\dfrac{2}{3}+\dfrac{5}{7}-\dfrac{3}{14}=\dfrac{28}{42}+\dfrac{30}{42}-\dfrac{9}{42}=\dfrac{49}{42}=\dfrac{7}{6}\)
d) \(\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{9}{45}=\dfrac{2}{5}-\dfrac{3}{7}+\dfrac{1}{5}=\dfrac{14}{35}-\dfrac{15}{35}+\dfrac{7}{35}=\dfrac{6}{35}\)
e) \(\dfrac{21}{47}+\dfrac{9}{45}+\dfrac{26}{47}+\dfrac{45}{5}=\dfrac{21}{47}+\dfrac{1}{5}+\dfrac{26}{47}+\dfrac{45}{5}=\left(\dfrac{21}{47}+\dfrac{26}{47}\right)+\left(\dfrac{1}{5}+\dfrac{45}{5}\right)\)
\(=1+\dfrac{46}{5}=\dfrac{51}{5}\)
f) \(\dfrac{15}{12}-\dfrac{18}{13}+\dfrac{5}{13}-\dfrac{3}{12}=\left(\dfrac{15}{12}-\dfrac{3}{12}\right)+\left(-\dfrac{18}{13}+\dfrac{5}{13}\right)=1+\left(-1\right)=0\)
g) \(\dfrac{-8}{18}-\dfrac{15}{27}=\dfrac{-4}{9}-\dfrac{5}{9}=\dfrac{-9}{9}=-1\)
h)\(\dfrac{3}{7}+\dfrac{-5}{2}-\dfrac{3}{5}=\dfrac{30}{70}+\dfrac{-175}{70}-\dfrac{42}{70}=\dfrac{-187}{70}\)
i) \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{11.16.3}{12.33.5}=\dfrac{4}{15}\)
a) \(\dfrac{37}{40}-0,64\\ =\dfrac{37}{40}-\dfrac{16}{25}\\ =\dfrac{185}{200}-\dfrac{128}{200}\\ =\dfrac{57}{200}\)
b) \(130\dfrac{25}{28}-120\dfrac{12}{35}\\ =\dfrac{3665}{28}-\dfrac{4212}{35}\\ =\dfrac{18325}{140}-\dfrac{16848}{140}\\ =\dfrac{211}{20}\)
A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)
A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)
A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)
A=\(\dfrac{7}{24}\)
B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)
B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)
B=\(1+\left(-1\right)+\left(-1\right)=-1\)
C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)
C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)
D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)
D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)
D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)
Tính các tổng dưới đây sau khi đã rút gọn phân số :
a)\(\dfrac{7}{21}\) + \(\dfrac{9}{-36}\) = \(\dfrac{7}{21}\)+\(\dfrac{-9}{36}\)=\(\dfrac{1}{3}\)+\(\dfrac{-1}{4}\)=\(\dfrac{4}{12}\)+\(\dfrac{-3}{12}\)=\(\dfrac{1}{12}\)
b) \(\dfrac{-12}{18}\)+\(\dfrac{-21}{35}\)=\(\dfrac{-2}{3}\)+\(\dfrac{-3}{5}\)=\(\dfrac{-10}{15}\)+\(\dfrac{-9}{15}\)=\(\dfrac{-19}{15}\)
c) \(\dfrac{-3}{21}\)+\(\dfrac{6}{42}\)=\(\dfrac{-1}{7}\)+\(\dfrac{1}{7}\)=0
d) \(\dfrac{-18}{24}\)+\(\dfrac{15}{-21}\)=\(\dfrac{-18}{24}\)+\(\dfrac{-15}{21}\)=\(\dfrac{-3}{4}\)+\(\dfrac{-5}{7}\)=\(\dfrac{-21}{28}\)+\(\dfrac{-20}{28}\)=\(\dfrac{-41}{28}\)
a: 2/9=4/18
1/3=6/18
5/18=5/18
b: 7/15=14/30
1/5=6/30
-5/6=-25/30
c: -21/56=-3/7
-3/16=-63/336
5/24=70/336
-21/56=-3/7=-144/336
d: \(\dfrac{-4}{7}=\dfrac{-36}{63}\)
8/9=56/63
\(-\dfrac{10}{21}=-\dfrac{30}{63}\)
e: 3/-20=-3/20=-9/60
-11/-30=11/30=22/60
7/15=28/60
\(=\dfrac{15}{7\cdot8}-\dfrac{13}{6\cdot7}+\dfrac{11}{5\cdot6}-\dfrac{9}{4\cdot5}+\dfrac{7}{3\cdot4}-\dfrac{5}{2\cdot3}+\dfrac{3}{1\cdot2}\)
\(=\dfrac{1}{8}+\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{2}\)
=1+1/8=9/8
a) ta co:
1/18<x/12<y/9<1/4
=>2/36<x.3/36<y.4/36<9/36
=>x.3thuộc{3;6};y.4thuộc{4;8}
=>x thuộc{1;2};y thuộc{1:2}
b) ta co
7/8<x/40<9/10
=>70/80<x.2/40<72/80
=>x.2 =71
=>x=71/2
\(a,\dfrac{-5}{21}+\dfrac{-6}{42}=\dfrac{-5}{21}+\dfrac{-3}{21}=\dfrac{\left(-5\right)+\left(-3\right)}{21}=\dfrac{-8}{21}\)
\(b,\dfrac{16}{40}+\dfrac{-27}{45}=\dfrac{2}{5}+\dfrac{-3}{5}=\dfrac{2+\left(-3\right)}{5}=\dfrac{-1}{5}\)
\(c,\dfrac{-4}{18}+\dfrac{-12}{30}=\dfrac{-2}{9}+\dfrac{-2}{5}=\dfrac{-10}{45}+\dfrac{-18}{45}=\dfrac{\left(-10\right)+\left(-18\right)}{45}=\dfrac{-28}{45}\)
a)\(\dfrac{-5}{21}\)+\(\dfrac{-6}{42}\)= \(\dfrac{-10}{42}\)+\(\dfrac{-6}{42}\)= \(\dfrac{-16}{42}\)= \(\dfrac{-8}{21}\)
b)\(\dfrac{16}{40}\)+ \(\dfrac{-27}{45}\)= \(\dfrac{2}{5}\)+\(\dfrac{-3}{5}\)= \(\dfrac{-5}{5}\)= (-1)
c) \(\dfrac{-4}{18}\)+\(\dfrac{-12}{30}\)= \(\dfrac{-2}{9}\)+\(\dfrac{-2}{5}\)= \(\dfrac{-10}{45}\)+\(\dfrac{-18}{45}\)= \(\dfrac{-28}{45}\)
\(\dfrac{18}{45}\) + \(\dfrac{16}{12}\) + \(\dfrac{6}{30}\) - \(\dfrac{6}{9}\) - \(\dfrac{21}{35}\)
= \(\dfrac{2}{5}\) + \(\dfrac{4}{3}\) + \(\dfrac{1}{5}\) - \(\dfrac{2}{3}\) - \(\dfrac{3}{5}\)
= (\(\dfrac{2}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{3}{5}\)) + (\(\dfrac{4}{3}\) - \(\dfrac{2}{3}\))
= (\(\dfrac{3}{5}\) - \(\dfrac{3}{5}\)) + \(\dfrac{2}{3}\)
= 0 + \(\dfrac{2}{3}\)
= \(\dfrac{2}{3}\)
18454518 + 16121216 + 630306 - 6996 - 21353521
= 2552 + 4334 + 1551 - 2332 - 3553
= (2552 + 1551 - 3553) + (4334 - 2332)
= (3553 - 3553) + 2332
= 0 + 2332
= 2332