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Bài 1:
a) \(\mathbb{Z}\)
b) \(\mathbb{Q}\)
c) \(\mathbb{Q}\)
d) \(\mathbb{Z}\)
e) \(\mathbb{N}\)
Bài 2:
a) Ta thấy: \(\frac{1}{8}>0; \frac{-3}{8}< 0\) \(\Rightarrow \frac{1}{8}> \frac{-3}{8}\)
b) \(\frac{-3}{7}< 0; 2\frac{1}{2}>0\Rightarrow \frac{-3}{7}< 2\frac{1}{2}\)
c) \(-3,9< 0< 0,1\)
d) \(-2,3< 0< 3,2\)
a) \(\frac{1}{8}>0>\frac{-3}{8}=>\frac{1}{8}>\frac{-3}{8}\)
b) \(\frac{-3}{7}< 0< 2\frac{1}{2}=>\frac{-3}{7}< 2\frac{1}{2}\)
c) \(-3.9< 0< 0.1=>-3.9< 0.1\)
d) \(-2.3< 0< 3.2=>-2.3< 3.2\)
\(-1\in Z;-1\in Q\\ \dfrac{7}{123}\in Q\\ 3,05\in Q\\ \dfrac{-2}{3}\in Q\\ 1035\in N;1035\in Z;1035\in Q\)
a) ta có : \(\dfrac{1}{8}>0\) và \(\dfrac{-3}{8}< 0\) \(\Rightarrow\dfrac{-3}{8}< \dfrac{1}{8}\) vậy \(\dfrac{-3}{8}< \dfrac{1}{8}\)
b) ta có : \(\dfrac{-3}{7}< 0\) và \(2\dfrac{1}{2}=\dfrac{5}{2}>0\) \(\Rightarrow\dfrac{-3}{7}< 1\dfrac{1}{2}\) vậy \(\dfrac{-3}{7}< 1\dfrac{1}{2}\)
c) ta có : \(-3,9< 0\) và \(0,1>0\) \(\Rightarrow-3,9< 0,1\) vậy \(-3,9< 0,1\)
d) ta có : \(-2,3< 0\) và \(3,2>0\) \(\Rightarrow-2,3< 3,2\) vậy \(-2,3< 3,2\)
a)\(\dfrac{1}{8}>0>\dfrac{-3}{8}\)
=>\(\dfrac{1}{8}>\dfrac{-3}{8}\)
b)\(2\dfrac{1}{2}=\dfrac{5}{2}>0>\dfrac{-3}{7}\)
=>\(2\dfrac{1}{2}>\dfrac{-3}{7}\)
c)\(-3,9< 0< 0,1\)
=>\(-3,9< 0,1\)
d)\(-2,3< 0< 3,2\)
=>\(-2,3< 3,2\)
a)1/8>-3/8 Vì cùng mẫu nên so sánh tử số mà 1>-3
b)-3/7<5/2
c) -3,9<0,1 Vì (-)<(+)
d)-2,3<0,1 Vì (-)<(+)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a: \(\dfrac{3.7}{4.5}=\dfrac{37}{45}\)
b: \(=\dfrac{4}{35}\cdot\dfrac{5}{2}=\dfrac{10}{70}=\dfrac{1}{7}\)
c: \(=\dfrac{7}{3}\cdot\dfrac{4}{3}=\dfrac{28}{9}\)
d: \(=\dfrac{22}{7}:\dfrac{5}{3}=\dfrac{22}{7}\cdot\dfrac{3}{5}=\dfrac{66}{35}\)
e: \(=2:\dfrac{1}{200}=\dfrac{400}{1}\)
f: =13/5:37/10=13/5x10/37=26/37
1
A)Z ; Q B)Q C)Q D)Q E)N ; Z ; Q
2
A)> B)< C)< D)<