Điền dấu < ; > ; =
3\(\dfrac{1}{5}-\dfrac{8}{3}...........\dfrac{17}{5}-2\dfrac{12}{5}\)
\(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{89}+\dfrac{1}{90}\) ..... \(\dfrac{5}{6}\)
điền dấu < = > vào chỗ chấm
a)\(\dfrac{2}{3}\)+\(\dfrac{1}{4}\)+\(\dfrac{5}{6}\)...\(\dfrac{7}{4}\)
b)\(\dfrac{4}{5}\)+\(\dfrac{2}{3}\)*2-\(\dfrac{1}{5}\)...\(\dfrac{1}{2}\)
c)\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)...\(\dfrac{15}{16}\)
d)\(\dfrac{1515:101}{2525:101}\)...\(\dfrac{3}{5}\)
Tính:
a) \(\dfrac{13}{14}\)-\(\dfrac{-7}{8}\)+\(\dfrac{-3}{2}\)
b) \(\dfrac{5}{17}\)+\(\dfrac{-15}{34}\).\(\dfrac{2}{5}\)
c) \(\dfrac{1}{5}\):\(\dfrac{1}{10}\)-\(\dfrac{1}{3}\).(\(\dfrac{6}{5}\)-\(\dfrac{2}{4}\))
d) \(\dfrac{-3}{4}\):(\(\dfrac{12}{-5}\)-\(\dfrac{-7}{10}\))
*Lưu ý: Không viết luôn kết quả, giải chi tiết.
\(a,\dfrac{13}{14}\cdot\dfrac{-7}{8}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-3}{2}\)
\(=-\dfrac{13}{16}+\dfrac{-24}{16}\)
\(=-\dfrac{37}{16}\)
\(b,\dfrac{5}{17}+\dfrac{-15}{34}\cdot\dfrac{2}{5}\)
\(=\dfrac{5}{17}+\dfrac{-3}{17}\)
\(=\dfrac{2}{17}\)
\(c,\dfrac{1}{5}:\dfrac{1}{10}-\dfrac{1}{3}\cdot\left(\dfrac{6}{5}-\dfrac{2}{4}\right)\)
\(=2-\dfrac{1}{3}\cdot\dfrac{7}{10}\)
\(=2-\dfrac{7}{30}\)
\(=\dfrac{53}{30}\)
\(d,\dfrac{-3}{4}:\left(\dfrac{12}{-5}-\dfrac{-7}{10}\right)\)
\(=\dfrac{-3}{4}:\dfrac{-17}{10}\)
\(=\dfrac{15}{34}\)
Điền dấu thích hợp (>,<,=) vào chỗ trống.
a,\(\dfrac{-6}{11}+\dfrac{5}{-11}....--1\) b.\(\dfrac{-5}{16}+\dfrac{-3}{16}...\dfrac{-1}{3}\)
c,\(\dfrac{2}{5}...\dfrac{3}{4}+\dfrac{-1}{6}\) d,\(\dfrac{5}{6}+\dfrac{-2}{3}...\dfrac{1}{12}+\dfrac{-4}{5}\)
Giúp mk nha :>>. Mk cảm ơn.
a. \(\dfrac{-6}{11}+\dfrac{5}{-11}< --1\)
b. \(\dfrac{-5}{16}+\dfrac{-3}{16}>-\dfrac{1}{3}\)
c. \(\dfrac{2}{5}>\dfrac{3}{4}+-\dfrac{1}{6}\)
d. \(\dfrac{5}{6}+\dfrac{-2}{3}>\dfrac{1}{12}+\dfrac{-4}{5}\)
tính thuận tiện:
\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\) \(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\) \(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)
\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)
mik sẽ chỉ tick 3 bn xong trước phải chi tiết rõ ràng
a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
\(\dfrac{4}{5}+\dfrac{2}{3}+\dfrac{1}{5}+\dfrac{1}{3}\)
\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\)
\(=\dfrac{5}{5}+\dfrac{3}{3}\)
\(=1+1\)
\(=2\)
============
\(\dfrac{17}{12}+\dfrac{29}{7}-\dfrac{8}{7}+\dfrac{7}{12}\)
\(=\left(\dfrac{17}{12}+\dfrac{7}{12}\right)+\left(\dfrac{29}{7}-\dfrac{8}{7}\right)\)
\(=\dfrac{24}{12}+\dfrac{21}{7}\)
\(=2+3\)
\(=5\)
====================
\(\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{2}{5}-\dfrac{1}{7}-\dfrac{2}{14}\)
\(=\dfrac{9}{15}+\dfrac{16}{7}+\dfrac{6}{15}-\dfrac{1}{7}-\dfrac{1}{7}\)
\(=\left(\dfrac{9}{15}+\dfrac{6}{15}\right)+\left(\dfrac{16}{7}-\dfrac{1}{7}-\dfrac{1}{7}\right)\)
\(=\dfrac{15}{15}+\dfrac{14}{7}\)
\(=1+2\)
\(=3\)
===============
\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{2}{5}+\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{5}{5}+\dfrac{3}{3}+\dfrac{8}{4}\)
\(=1+1+2\)
\(=4\)
Điền dấu > ; < hoặc = vào chỗ chấm cho thích hợp:
a) \(\dfrac{3}{4}.....\dfrac{4}{5}\) b) \(\dfrac{11}{12}.....\dfrac{7}{8}\) c) \(\dfrac{5}{7}+\dfrac{3}{5}.....\dfrac{6}{7}:\dfrac{5}{8}\) d) \(\dfrac{9}{16}-\dfrac{5}{9}.....\dfrac{1}{16}x\dfrac{1}{9}\)
tính một cách hợp lí:
a) \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
b) \(\left(\dfrac{-1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(\dfrac{-15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
c) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
giải chi tiết giúp mình nha
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: =−2590+6490−8190=−2590+6490−8190
(−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−53+4229=−53+4229
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
BT5 : Chứng minh
2) \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\)
Giải:
Đặt \(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)
Ta có:
\(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)
\(\Rightarrow A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)
Nhận xét:
\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)
\(\Rightarrow A< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)
\(\Rightarrow A< \dfrac{4}{5}\left(1\right)\)
Lại có:
\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{1}{6}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\)
\(\Rightarrow A>\dfrac{3}{5}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{3}{5}< A< \dfrac{4}{5}\)
Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)
Đặt A=131+132+133+...+159+160A=131+132+133+...+159+160
Ta có:
A=131+132+133+...+159+160A=131+132+133+...+159+160
⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)
Nhận xét:
131+132+...+140<130+130+...+130=13131+132+...+140<130+130+...+130=13
141+142+...+150<140+140+...+140=14141+142+...+150<140+140+...+140=14
151+152+...+160<150+150+...+150=15151+152+...+160<150+150+...+150=15
⇒A<13+14+15=4760<4860=45⇒A<13+14+15=4760<4860=45
⇒A<45(1)⇒A<45(1)
Lại có:
131+132+...+140>140+140+...+140=14131+132+...+140>140+140+...+140=14
141+142+...+150>150+150+...+150=15141+142+...+150>150+150+...+150=15
151+152+...+160>160+160+...+160=16151+152+...+160>160+160+...+160=16
⇒A>14+15+16=3760>3660=35⇒A>14+15+16=3760>3660=35
⇒A>35(2)⇒A>35(2)
Từ (1)(1) và (2)(2)
⇒35<A<45⇒35<A<45
Vậy 35<131+132+133+...+159+160<4535<131+132+133+...+159+160<45
Bài 1: tính
Cho A= \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+........+\dfrac{1}{60}>\dfrac{7}{12}\)
B=\(\dfrac{1}{3^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}+.....+\dfrac{1}{50^2}\)
CMR B > \(\dfrac{1}{4}\); B < \(\dfrac{4}{9}\)
C = \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}.\dfrac{7}{8}...........\dfrac{79}{80}\)<\(\dfrac{1}{9}\)
a)\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{7}_{ }+\dfrac{5}{-8}\)
b)\(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}_{ }+\dfrac{-8}{13}\)
c)\(\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
d)\(\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
e)\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)
f)\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
Các bạn không cần trả lời câu hỏi trên của mik vì mik đã hiểu rồi nha . Cho nên đừng trả lời ! OK