K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

7 tháng 10 2023

\(\dfrac{4}{7}-\dfrac{1}{7}.\dfrac{10}{13}\)

\(=\dfrac{4}{7}-\dfrac{10}{91}\)

\(=\dfrac{52}{91}-\dfrac{10}{91}\)

\(=\dfrac{42}{91}=\dfrac{6}{13}\)

_______

\(\dfrac{1}{8}+\dfrac{1}{6}:\dfrac{2}{3}\)

\(=\dfrac{1}{8}+\dfrac{1}{6}.\dfrac{3}{2}\)

\(=\dfrac{1}{8}+\dfrac{1}{4}\)

\(=\dfrac{1}{8}+\dfrac{2}{8}\)

\(=\dfrac{3}{8}\)

___________

\(\left(\dfrac{7}{9}-\dfrac{2}{18}\right):\dfrac{4}{7}\)

\(=\left(\dfrac{14}{18}-\dfrac{2}{18}\right).\dfrac{7}{4}\)

\(=\dfrac{12}{18}.\dfrac{7}{4}\)

\(=\dfrac{7}{6}\)

\(#WendyDang\)

AH
Akai Haruma
Giáo viên
7 tháng 10 2023

Lời giải:

$\frac{4}{7}-\frac{1}{7}\times \frac{10}{13}=\frac{1}{7}\times 4-\frac{1}{7}\times \frac{10}{13}$

$=\frac{1}{7}\times (4-\frac{10}{13})=\frac{1}{7}\times \frac{42}{13}$
$=\frac{42}{7\times 13}=\frac{6\times 7}{7\times 13}=\frac{6}{13}$
-----------------

$\frac{1}{8}+\frac{1}{6}: \frac{2}{3}=\frac{1}{8}+\frac{1}{6}\times \frac{3}{2}=\frac{1}{8}+\frac{1}{4}=\frac{1}{8}+\frac{2}{8}=\frac{3}{8}$

------------------

$(\frac{7}{9}-\frac{2}{18}): \frac{4}{7}=(\frac{7}{9}-\frac{1}{9})\times \frac{7}{4}$
$=\frac{6}{9}\times \frac{7}{4}=\frac{2}{3}\times \frac{7}{4}=\frac{7}{6}$

1: =72/90+65/90=137/90

2: =24/56-77/56=-53/56

3: =-7/10+4/5=1/10

4: =15/100-4/100=11/100

5: =4/6-5/6=-1/6

6: =10/40-15/40-76/40=-81/40

7: =-9/10+7/18

=-81/90+35/90=-46/90=-23/45

8: =27/90-55/90=-28/90=-14/45

9: =36/60-50/60-35/60=-49/60

10: =-4/9+5/6-3/8

=-32/72+60/72-27/72

=1/72

10 tháng 1 2022

\(1.\) \(\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}+\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{7+32-24}{36}=\dfrac{5}{12}.\)

\(2.\) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}=\dfrac{-9-2-7}{18}+\dfrac{4+3}{7}=\dfrac{-18}{18}+\dfrac{7}{7}=-1+1=0.\)

\(3.\) \(-\dfrac{10}{3}+\dfrac{13}{10}-\dfrac{1}{6}+\dfrac{1}{10}=\dfrac{13+1}{10}+\dfrac{-20-1}{6}=\dfrac{14}{10}+\dfrac{-21}{6}=\dfrac{7}{5}-\dfrac{7}{2}=-\dfrac{21}{10}.\)

\(4.\) \(\dfrac{10}{17}-\dfrac{5}{13}-\left(-\dfrac{7}{17}\right)-\dfrac{8}{13}+\dfrac{11}{25}=\dfrac{10+7}{17}+\dfrac{-5-8}{13}+\dfrac{11}{25}=\dfrac{17}{17}-\dfrac{13}{13}+\dfrac{11}{25}=\dfrac{11}{25}.\)

1: \(=\dfrac{7}{36}+\dfrac{32}{36}+\dfrac{-24}{36}=\dfrac{15}{36}=\dfrac{5}{12}\)

2: \(=\left(-\dfrac{1}{2}-\dfrac{1}{9}-\dfrac{7}{18}\right)+\dfrac{3}{7}+\dfrac{4}{7}\)

\(=\dfrac{-9-2-7}{18}+1=-1+1=0\)

3: \(=\left(-\dfrac{10}{3}-\dfrac{1}{6}\right)+\dfrac{14}{10}\)

\(=-\dfrac{21}{6}+\dfrac{14}{10}=\dfrac{-7}{2}+\dfrac{14}{10}=\dfrac{-35+14}{10}=-\dfrac{21}{10}\)

4: \(=\left(\dfrac{10}{17}+\dfrac{7}{17}\right)-\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\dfrac{11}{25}=\dfrac{11}{25}\)

10 tháng 8 2022

 

1.3.77−1​+3.7.99−3​+7.9.1313−7​+9.13.1515−9​+\frac{19-13}{13.15.19}+13.15.1919−13​

=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}=1.31​−3.71​+3.71​−7.91​+7.91​−9.131​+9.131​−13.151​+13.151​−15.191​

=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}=1.31​−15.191​=28595​−2851​=28594​

b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)b,=61​.(1.3.76​+3.7.96​+7.9.136​+9.13.156​+13.15.196​)

làm giống như trên

c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)c,=81​.(1.2.31​+2.3.41​+3.4.51​+...+48.49.501​)

=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)=161​.(1.2.32​+2.3.42​+3.4.52​+...+48.49.502​)

=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)=161​.(1.2.33−1​+2.3.44−2​+3.4.55−3​+...+48.49.5050−48​)

=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)=161​.(1.21​−2.31​+2.31​−3.41​+3.41​−4.51​+...+48.491​−49.501​)

=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}=161​.(21​−24501​)=161​.(24501225​−24501​)=4900153​

d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)d,=75​.(1.5.87​+5.8.127​+8.12.157​+...+33.36.407​)

=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)=75​.(1.5.88−1​+5.8.1212−5​+8.12.1515−8​+...+33.36.4040−33​)

=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)=75​.(1.51​−5.81​+5.81​−8.121​+8.121​−12.151​+...+33.361​−36.401​)

=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}=75​.(51​−14401​)=75​.(1440288​−14401​)=28841​

P/S: . là nhân nha

18 tháng 4 2020

Ghi đầy đủ nha

6 tháng 3 2022

bn có thể ghi rõ ràng đc ko?

18 tháng 4 2023

\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)

\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)

\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)

\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)

\(=-\dfrac{38}{42}\)

\(=-\dfrac{19}{21}\)

\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)

\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)

\(=\dfrac{13}{13}+11\)

\(=1+11\)

\(=12\)

\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)

\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)

\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)

\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)

\(=\dfrac{28}{7}-\dfrac{31}{9}\)

\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)

\(=\dfrac{252}{63}-\dfrac{217}{63}\)

\(=\dfrac{35}{63}\)

\(=\dfrac{5}{9}\)

\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)

\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1}{2}\)

\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)

\(=\dfrac{-15}{24}+\dfrac{12}{24}\)

\(=\dfrac{-3}{24}\)

\(=-\dfrac{1}{8}\)

\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)

\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)

\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)

\(=-1+1-\dfrac{3}{7}\)

\(=-\dfrac{3}{7}\)

\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)

\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)

\(=\dfrac{6}{5}\times\dfrac{20}{7}\)

\(=\dfrac{120}{35}\)

\(=\dfrac{24}{7}\)

 

a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)

\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)

\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)

=0

b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)

\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)

\(=\dfrac{7}{3}\cdot0=0\)

c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)

\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)

d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)

\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)

24 tháng 1

\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)

\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)

\(=0:\dfrac{3}{8}=0\)

\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)

\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)

\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)

\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)

\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)

\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)

\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)

20 tháng 8 2017

\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)

\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)

\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)

\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)

làm giống như trên

\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)

\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)

P/S: . là nhân nha

sxasxsxxsxsxssxsxsxsxsx232332321322