b) Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời. 8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\) =\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\) =\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\) =7.(-7) =-49 Ta có: \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) \(\Rightarrow ad+cd< bc+dc\) \(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\) \(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\) (1) \(ad< bc\) \(\Rightarrow ad+ab< bc+ab\) \(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\) \(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\) (2) Từ (1), (2) \(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\left(đpcm\right)\) Ta có : \(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) \(\Rightarrow ad+ab< bc+ab\Rightarrow a\left(d+b\right)< b\left(c+a\right)\) \(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\) Lại có : \(ad< bc\Rightarrow ad+cd< bc+cd\Rightarrow d\left(a+c\right)< c\left(b+d\right)\) \(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\) Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\rightarrowđpcm\) Ta có : a, 25/7 + 13/21 - 11/7 + 17/21 + 1/3 . = ( 25/7 - 11/7 ) + ( 13/21 + 17/21 + 1/3 ) . = 2 + ( 20/21 + 7/21 ) . = 2 + 9/7 . = 23/7 . b, ( 1/3 + 12/67 + 13/41 ) - ( 79/67 - 28/41 ) . = 1/3 + 12/67 + 13/41 - 79/67 + 28/41 . = 1/3 + ( 12/67 - 79/67 ) + ( 13/41 + 28/41 ) . = 1/3 - 1 + 1 . = 1/3 . c, ( 11/4 . -5/9 - 4/9 . 11/4 ) . 8/33 . = [ 11/4 . ( -5/9 - 4/9 ) ] . 8/33 . = [ 11/4 . ( - 1 ) ] . 8/33 . = -11/4 . 8/33 . = -2/3 . d, 38/45 - ( 8/45 - 17/51 - 3/11 ) . = 38/45 - 8/45 + 17/51 + 3/11 . = 2/3 + 17/51 + 3/11 . = 374/561 + 187/561 + 153/561 . = 14/11 . A)0,25:(10,3-9,8)-3/4 =1/4:(103/10-49/5)-3/4 =1/4:1/2-3/4 =1/2-3/4 =2/4-3/4 =-1/4 B)-5/9.13/28-13/28.4/9 =-5/9-4/9.13/28 =-1.13/28 =-13/28 c)6/7+5/8:5-3/16 =6/7+1/8-3/16 =55/56-3/16 =89/112 d)-5/7.2/11+-5/7.9/11+1/5/7 =-5/7.(2/11+9/11)+12/7 =-5/7.1+12/7 =-5/7+12/7 =1 e)-7/12-8/15+11/20 =-67/60+11/20 =-17/30 f)-17/25.20/33+-17/25.13/33+-3/25 =-17/25.(20/33+13/33)-3/25 =-17/25.1-3/25 =-17/25-3/25 =-4/5 CHÚC BẠN HỌC TỐT............... NẾU ĐÚNG THÌ TICK CHO MK VỚI NHA HELLO HELLO.......... a: \(A=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{-5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}\) \(=\dfrac{-3}{5}+\dfrac{3}{5}=0\) b: \(=3^4-\left(-8\right)^2-\left(-25\right)^2\) \(=81-64-625=-608\) c: \(=2^3+3\cdot1\cdot\dfrac{1}{4}\cdot4+\left[4:\dfrac{1}{2}\right]:8\) \(=8+3+4\cdot2:8=11+1=12\) a: \(=\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\cdot\dfrac{18}{5}-\dfrac{6}{5}:\dfrac{-9}{5}+4\) \(=\dfrac{18}{5}-\dfrac{6}{5}\cdot\dfrac{-5}{9}+4\) \(=\dfrac{18}{5}+\dfrac{2}{3}+4\) \(=\dfrac{124}{15}\) b: \(=\dfrac{9}{25}\cdot\left(\dfrac{3}{5}-\dfrac{1}{5}+\dfrac{1}{2}\right)-\dfrac{3}{8}:\dfrac{9}{8}\) \(=\dfrac{9}{25}\cdot\dfrac{4}{10}-\dfrac{1}{3}\) \(=-\dfrac{71}{375}\) c: \(=\dfrac{7}{10}:\dfrac{4}{5}+\dfrac{2}{9}:\dfrac{5}{9}+\dfrac{1}{8}\) \(=\dfrac{7}{10}\cdot\dfrac{5}{4}+\dfrac{2}{5}+\dfrac{1}{8}\) =1+2/5 =7/5 d: \(=\dfrac{3}{7}\left(19+\dfrac{1}{3}-33-\dfrac{1}{3}\right)-\dfrac{2}{7}=\dfrac{3}{7}\cdot\left(-14\right)-\dfrac{2}{7}=-6-\dfrac{2}{7}=\dfrac{-44}{7}\) e: \(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{-2^{11}\cdot3^{11}-2^{12}\cdot3^{12}}\) \(=\dfrac{2^{12}\cdot3^{10}\cdot6}{-2^{11}\cdot3^{11}\left(1+2\cdot3\right)}=-\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{-4}{7}\) a, \(\left(\dfrac{-5}{11}\right).\dfrac{7}{15}.\left(\dfrac{11}{-5}\right).\left(-30\right)=\dfrac{-5}{11}.\dfrac{7}{15}.\dfrac{-11}{5}.\dfrac{-30}{1}\)= ( - 14 ) b, \(\left(\dfrac{11}{12}:\dfrac{33}{16}\right).\dfrac{3}{5}=\dfrac{11}{12}.\dfrac{16}{33}.\dfrac{3}{5}=\dfrac{1.4.3}{3.3.5}=\dfrac{4}{15}\) c, \(\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{15}{-7}.\left(-16\right)=\dfrac{-7}{15}.\dfrac{5}{8}.\dfrac{-15}{7}.\dfrac{-16}{1}\) \(\dfrac{-1.5.-1.-2}{1.1.1.1}=\left(-10\right)\) d,\(\left(\dfrac{-1}{2}\right).3\dfrac{1}{5}+\left(\dfrac{-1}{2}\right).-2\dfrac{1}{5}=\left(\dfrac{-1}{2}\right).\left[\dfrac{16}{5}+\left(\dfrac{-11}{5}\right)\right]\) = \(\left(\dfrac{-1}{2}\right).1=\dfrac{-1}{2}\) a) (-5/11.11/-5).7/15 =1.7/15=7/15 b)(11/12:33/16).3/5 =(11/12.16/33).3/5 =4/9.3/5=4/15 c)(-7/15.15/-7).5/8 =1.5/8=5/8 d)(-1/2).(16/5.-11/5) =-1/2.1=-1/2 xg r đó \(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\) \(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\) \(=\dfrac{18+12+\left(-5\right)}{24}\) \(=\dfrac{25}{24}\) \(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\) \(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\) \(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\) \(=\dfrac{5}{7}.0=0\) \(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\) \(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\) \(=1+1+\dfrac{1}{2}\) \(=2\dfrac{1}{2}\) \(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\) \(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\) \(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\) \(=\dfrac{1391}{714}\) a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\) b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\) c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\) d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\) a/Ta có: \(\dfrac{4}{3}-\left[\left(\dfrac{-11}{6}\right)-\left(\dfrac{2}{9}+\dfrac{5}{3}\right)\right]\) \(=\) \(\dfrac{4}{3}-\left[\dfrac{-11}{6}-\dfrac{2}{9}-\dfrac{5}{3}\right]\) \(=\) \(\dfrac{4}{3}+\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{3}\) \(=\) \(\dfrac{24}{18}+\dfrac{33}{18}+\dfrac{4}{18}+\dfrac{30}{18}\) \(=\) \(\dfrac{91}{18}\) b/Ta có: \(\left(8-\dfrac{9}{4}+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(3+\dfrac{2}{4}-\dfrac{9}{7}\right)\) \(=\) \(8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\) \(=\) \(8+6-3-\dfrac{9}{4}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\) \(=\) \(11-\dfrac{2}{4}+\dfrac{14}{7}\) \(=\) \(11-\dfrac{1}{2}+2\) \(=\) \(9-\dfrac{1}{2}\) \(=\) \(\dfrac{17}{2}\) Chúc bn học tốt!!! 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) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\) \(=1-\dfrac{1}{10}\) \(=\dfrac{9}{10}\)
K
Khách
NL
Nguyễn Lê Phước Thịnh
12 tháng 7 2023
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