\(\left(-\dfrac{1}{2}\right)^2\div\dfrac{1}{4}-2\times\left(-\dfrac{1}{2}\right)^2\\= \dfrac{1}{4}\div\dfrac{1}{4}-2\times\dfrac{1}{4}\\ =1-\dfrac{1}{2}\\ =\dfrac{1}{2}\)

\(\left(-2\right)^3\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right)\div\dfrac{5}{12}\)

=  \(-6\times-\dfrac{1}{24}+\left(\dfrac{4}{3}-\dfrac{11}{6}\right)\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{1}{2}\div\dfrac{5}{12}\)

=  \(\dfrac{1}{4}+-\dfrac{6}{5}\)

=  \(\dfrac{1}{4}-\dfrac{6}{5}\)

=  \(-\dfrac{19}{20}\)

\(\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\\ =\dfrac{58}{9}+\dfrac{7}{11}-\dfrac{40}{9}+\dfrac{26}{11}\\ =\dfrac{58}{9}-\dfrac{40}{9}+\dfrac{7}{11}+\dfrac{26}{11}\\ =12+3\\ =15\)

\(a,\left(\dfrac{-1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)

\(=\left(-\dfrac{1}{2}\right)^2\left(4-2\right)\)

\(=\dfrac{1}{4}.2=\dfrac{1}{2}\)

\(b,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)

\(=\left(-8\right).\dfrac{-1}{24}+\left(-\dfrac{1}{2}\right).\dfrac{12}{5}\)

\(=\dfrac{1}{3}+\left(-\dfrac{1}{5}\right)=\dfrac{2}{15}\)

\(c,\left(6\dfrac{4}{9}+\dfrac{7}{11}\right)-\left(4\dfrac{4}{9}-2\dfrac{4}{11}\right)\)

\(=\dfrac{701}{99}-\dfrac{206}{99}=\dfrac{495}{99}=5\)

\(d,10\dfrac{1}{5}-5\dfrac{1}{2}.\dfrac{60}{11}+\dfrac{3}{15\%}\)

\(=\dfrac{51}{5}-30+20=\dfrac{1}{5}\)

\(e,\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)

\(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}.\left(-\dfrac{7}{11}\right)\)

\(=-\dfrac{5}{11}\)

\(f,\dfrac{-5}{7}.\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)

\(=\left(-\dfrac{5}{7}\right)\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\)

\(=\left(-\dfrac{5}{7}\right)+\dfrac{12}{7}=1\)

\(10\dfrac{1}{5}-5\dfrac{1}{2}\times\dfrac{60}{11}+3\div15\%\\ =\dfrac{51}{5}-\dfrac{11}{2}\times\dfrac{60}{11}+3\div\dfrac{15}{100}\\ =\dfrac{51}{5}-30+20\\ =10,2-30+20\\ =0,2\)

\(\dfrac{5}{7}\times\dfrac{5}{11}+\dfrac{5}{7}\times\dfrac{2}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times-\dfrac{7}{11}=-\dfrac{5}{11}\)

\(-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+1\dfrac{5}{7}\\ =-\dfrac{5}{7}\times\dfrac{2}{11}+-\dfrac{5}{7}\times\dfrac{9}{11}+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ =-\dfrac{5}{7}\times1+\dfrac{12}{7}\\ =1\)

a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)

= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)

= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)

= \(1-\frac{1}{10}\)

=\(\frac{9}{10}\)

b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

=\(1-\frac{1}{11}\)

= \(\frac{10}{11}\)

c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

     \(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

     \(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

      \(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)

     \(\frac{2}{3}A\)=\(\frac{10}{11}\)

         A= \(\frac{10}{11}:\frac{2}{3}\)

          A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)

d) giả tương tự câu c kết quả \(\frac{25}{11}\)

tổng đặc biệt đó bạn

\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(1-\frac{1}{10}=\frac{9}{10}\)

những câu sau cũng áp dụng như vậy nhé

\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\)

\(=1-\frac{1}{11}=\frac{10}{11}\)

\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)

\(=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{3}{2}.\left(1-\frac{1}{11}\right)=\frac{3}{2}.\frac{10}{11}=\frac{15}{11}\)

Bài 1:

a: \(x=\dfrac{2}{3}:\dfrac{3}{5}=\dfrac{2}{3}\cdot\dfrac{5}{3}=\dfrac{10}{9}\)

b: \(x=\dfrac{17}{8}:\dfrac{7}{17}=\dfrac{17}{8}\cdot\dfrac{17}{7}=\dfrac{289}{56}\)

c: \(x=-\dfrac{3}{4}:\dfrac{7}{12}=\dfrac{-3}{4}\cdot\dfrac{12}{7}=\dfrac{-63}{28}=-\dfrac{9}{4}\)

d: \(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{1}{4}\)

hay \(x=\dfrac{1}{4}:\dfrac{1}{6}=\dfrac{3}{2}\)

e: \(\Leftrightarrow\dfrac{1}{2}:x=-4-\dfrac{1}{3}=-\dfrac{17}{3}\)

hay \(x=-\dfrac{1}{2}:\dfrac{17}{3}=\dfrac{-3}{34}\)

dad

a. 

\(\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{12}\right)+\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{12}\right)\) 

\(=\left(\frac{5}{4}+\frac{3}{5}+\frac{3}{8}-\frac{7}{5}\right):\left(-\frac{11}{12}\right)\)  

\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{12}\right)\) 

\(=\frac{33}{40}:\left(-\frac{11}{12}\right)\) 

\(=\frac{33}{40}\cdot\left(-\frac{12}{11}\right)\) 

\(=\frac{-9}{10}\)  

b. 

\(\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{15}\right)+\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\) 

\(=\left(\frac{3}{8}-\frac{7}{5}+\frac{5}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)  

\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{15}\right)\) 

\(=\frac{33}{40}:\left(-\frac{11}{15}\right)\) 

\(=\frac{33}{40}\cdot\left(-\frac{15}{11}\right)\) 

\(=\frac{-9}{8}\)

A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)

A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)

A=3.(1-1/400)

A=3.399/400

A=1197/400

A=3.(1/1.2+1/2.3+1/3.4+.....+1/399.400)

A=3.(1/1-1/2+1/2-1/3+......+1/399-1/400)

A=3.(1-1/400)

A=3.399/400

A=1197/400

Bài 1: Tính nhanh:

A = 3/1*2 + 3/2*3 + 3/3*4 + ... + 3/399*400

=>3A = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/399*400

    3A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/399 - 1/400

    3A = 1 - 1/400

      3A = 400/400 - 1/400

      3A = 399/400

        A = 399/400 : 3

        A = 399/400 . 1/3

        A = 133/400.

Có gì ko hiểu bn ib mk nha.^^

\(A=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{399.400}\)

\(A=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)

\(A=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)

\(A=3.\left(1-\frac{1}{400}\right)\)

\(A=3.\frac{399}{400}\)

\(A=\frac{1197}{400}\)

\(B=\frac{5}{1.2}+\frac{5}{2.3}+\frac{5}{3.4}+...+\frac{5}{399.400}\)

\(B=5.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{399.400}\right)\)

\(B=5.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{399}-\frac{1}{400}\right)\)

\(B=5.\left(1-\frac{1}{400}\right)\)

\(B=5.\frac{399}{400}\)

\(B=\frac{399}{80}\)

\(C=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\)

\(C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\)

\(C=\frac{1}{5}-\frac{1}{151}\)

\(C=\frac{146}{755}\)

\(D=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{149.151}\)

\(D=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{149.151}\right)\)

\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{149}-\frac{1}{151}\right)\)

\(D=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{151}\right)\)

\(D=\frac{3}{2}.\frac{146}{755}\)

\(D=\frac{219}{755}\)

\(E=\frac{11}{1.3}+\frac{11}{3.5}+\frac{11}{5.7}+...+\frac{11}{99.101}\)

\(E=\frac{11}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)

\(E=\frac{11}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)

\(E=\frac{11}{2}.\left(1-\frac{1}{101}\right)\)

\(E=\frac{11}{2}.\frac{100}{101}\)

\(E=\frac{550}{101}\)

_Chúc bạn học tốt_

Xóa nhanh đi ko CTV xóa giờ

Chia chúng nó ra

Câu dễ vậy bạn cần hỏi

a)  \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{20}\right)\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{19}{20}=\frac{1}{20}\)

b)  \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)

=>  \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

=> \(2A-A=\left(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

=>  \(A=2-\frac{1}{2^{2012}}\)

c) \(\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)

\(=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)

\(=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{231}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\)

\(=\frac{231}{4}.\frac{4}{21}=11\)

d.e)  ktra lại đề

a: \(=\dfrac{5}{7}\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)=\dfrac{5}{7}\cdot\dfrac{-7}{11}=-\dfrac{5}{11}\)

b: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\left(4+\dfrac{3}{8}\right)\)

\(=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)

c: \(=\dfrac{2}{15}-\dfrac{2}{15}\cdot5+\dfrac{3}{15}=\dfrac{2}{15}\cdot\left(-4\right)+\dfrac{3}{15}=\dfrac{-8+3}{15}=\dfrac{-5}{15}=-\dfrac{1}{3}\)

d: \(=\dfrac{4}{9}\left(19+\dfrac{1}{3}-39-\dfrac{1}{3}\right)=\dfrac{4}{9}\cdot\left(-20\right)=-\dfrac{80}{9}\)

`4/7+4`

`=4/7+4/1`

`=4/7+28/7`

`=32/7`

__

`3+6/11`

`=33/11+6/11`

`=39/11`

__

`3-5/7`

`=3/1-5/7`

`=21/7-5/7`

`=16/7`

__

`21/9-2`

`=21/9-18/9`

`=3/9`

`=1/3`

__

`15/24+2`

`=15/24+48/24`

`=63/24`

`=21/16`

__

`63/45-20/25`

`=63/45-4/5`

`=63/45-36/45`

`=27/45`

`=9/15`

__

`3/4-2/8`

`=3/4-1/4`

`=2/4`

__

`6/7-5/8`

`=48/56-35/56`

`=13/56`

__

`37/45-5/9`

`=37/45-25/45`

`=12/45`

`=4/15`

__

`46/39-11/13`

`=46/39-33/39`

`=13/39`

`=1/2`

__

`5/12+3/4+1/3`

`=5/12+9/12+4/12`

`=14/12+4/12`

`=18/12`

`=3/2`

__

`1/2+3/7+11/14`

`=7/14+6/14+11/14`

`=13/14+11/14`

`=24/14`

`=12/7`

__

`7/10-(1/5+1/4)`

`=7/10-(4/20+5/20)`

`=7/10-9/20`

`=14/20-9/20`

`=5/20`

`=1/4`

__

`15/4-2/3-3/4`

`=(15/4-3/4)-2/3`

`=12/4-2/3`

`=3-2/3`

`=9/3-2/3`

`=7/3`