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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) A= \(\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7}\)
= \(\left(\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}\right)+\left(\dfrac{3}{4}+\dfrac{5}{4}\right)\)
= \(\dfrac{4+2+1}{7}+\dfrac{3+5}{4}\)
= \(\dfrac{7}{7}+\dfrac{8}{4}\) = \(1+2\) = \(3\)
b) B= \(\dfrac{-4}{12}+\dfrac{18}{45}+\dfrac{-6}{9}+\dfrac{-21}{35}+\dfrac{6}{30}\)
= \(\dfrac{-1}{3}+\dfrac{2}{5}+\dfrac{-2}{3}+\dfrac{-3}{5}+\dfrac{1}{5}\)
= \(\left(\dfrac{-1}{3}+\dfrac{-2}{3}\right)+\left(\dfrac{2}{5}+\dfrac{-3}{5}+\dfrac{1}{5}\right)\)
= \(\dfrac{\left(-1\right)+\left(-2\right)}{3}+\dfrac{2+\left(-3\right)+1}{5}\)
= \(\dfrac{-3}{3}+\dfrac{0}{5}\) = \(-1+0\) = \(-1\)
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\(a,\dfrac{-5}{21}+\dfrac{-6}{42}=\dfrac{-5}{21}+\dfrac{-3}{21}=\dfrac{\left(-5\right)+\left(-3\right)}{21}=\dfrac{-8}{21}\)
\(b,\dfrac{16}{40}+\dfrac{-27}{45}=\dfrac{2}{5}+\dfrac{-3}{5}=\dfrac{2+\left(-3\right)}{5}=\dfrac{-1}{5}\)
\(c,\dfrac{-4}{18}+\dfrac{-12}{30}=\dfrac{-2}{9}+\dfrac{-2}{5}=\dfrac{-10}{45}+\dfrac{-18}{45}=\dfrac{\left(-10\right)+\left(-18\right)}{45}=\dfrac{-28}{45}\)
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) \(\dfrac{-4}{18}\)+\(\dfrac{-12}{30}\)= \(\dfrac{-2}{9}\)+\(\dfrac{-2}{5}\)= \(\dfrac{-10}{45}\)+\(\dfrac{-18}{45}\)= \(\dfrac{-28}{45}\)
a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)
\(\Rightarrow-4< x< \dfrac{-3}{10}\)
\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)
\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)
b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)
\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)
\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)
\(\Rightarrow x=\varnothing\)
c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)
\(\Rightarrow x\in\left\{1;2\right\}\)
+) Với \(x=1\)
\(\Rightarrow y\in\left\{1;2\right\}\)
+) Với \(x=2\)
\(\Rightarrow y=2\)
Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).
1) \(\dfrac{5}{6}-\dfrac{6}{7}+\dfrac{7}{8}-\dfrac{8}{9}+\dfrac{10}{9}-\dfrac{5}{6}+\dfrac{6}{7}-\dfrac{7}{8}+\dfrac{8}{9}\)
\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)-\left(\dfrac{6}{7}+\dfrac{6}{7}\right)+\left(\dfrac{7}{8}-\dfrac{7}{8}\right)-\left(\dfrac{8}{9}+\dfrac{8}{9}\right)+\dfrac{10}{9}\)
\(=0-0+0-0+\dfrac{10}{9}\)
\(=\dfrac{10}{9}\)
2) \(\dfrac{1}{13}+\dfrac{16}{7}+\dfrac{3}{105}-\dfrac{9}{7}-\dfrac{-12}{13}\)
\(=\left(\dfrac{1}{13}-\left(-\dfrac{12}{13}\right)\right)+\left(\dfrac{16}{7}-\dfrac{9}{7}\right)+\dfrac{3}{105}\)
\(=1+1+\dfrac{3}{105}\)
\(=\dfrac{213}{105}=\dfrac{71}{35}\)
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) 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
\(D=\dfrac{1}{2}+\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{1}{6}+\dfrac{-3}{35}+\dfrac{1}{3}+\dfrac{1}{41}\)
\(D=\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{3}\right)+\left(\dfrac{-1}{5}+\dfrac{-5}{7}+\dfrac{-3}{35}\right)+\dfrac{1}{41}\)
\(D=1+-1+\dfrac{1}{41}\)
\(D=0+\dfrac{1}{41}\)
\(D=\dfrac{1}{41}\)
\(C=\left(\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}\right)+\left(\dfrac{-3}{4}+\dfrac{-1}{36}+\dfrac{-2}{9}\right)+\dfrac{1}{57}\)
\(=\dfrac{5+9+1}{15}+\dfrac{-27-1-8}{36}+\dfrac{1}{57}\)
=1/57
\(E=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\left(\dfrac{3}{5}+\dfrac{4}{35}+\dfrac{2}{7}\right)+\dfrac{1}{127}=\dfrac{1}{127}\)
\(\dfrac{18}{45}\) + \(\dfrac{16}{12}\) + \(\dfrac{6}{30}\) - \(\dfrac{6}{9}\) - \(\dfrac{21}{35}\)
= \(\dfrac{2}{5}\) + \(\dfrac{4}{3}\) + \(\dfrac{1}{5}\) - \(\dfrac{2}{3}\) - \(\dfrac{3}{5}\)
= (\(\dfrac{2}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{3}{5}\)) + (\(\dfrac{4}{3}\) - \(\dfrac{2}{3}\))
= (\(\dfrac{3}{5}\) - \(\dfrac{3}{5}\)) + \(\dfrac{2}{3}\)
= 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