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b) -3.7^4 + 7^3 / 7^5 . 6 - 7^3 .2
= -21 . 7^3 + 7^3 . 1 / 7^3 . 294 - 7^3 . 2
= 7^3 . (-21+1) / 7^3 . (294 - 2)
= 20/292 = 5/73

3 tháng 10 2021

Tui Hông Hỉu, Dải Đáp Cho Tui Dới.............................

21 tháng 8 2016

\(\frac{28}{14}=2\)

\(\frac{5}{2}:2=\frac{5}{4}\)

\(\frac{8}{4}=2\)

\(\frac{1}{2}:\frac{2}{3}=\frac{3}{4}\)

\(\frac{3}{10}\)

\(\frac{21}{10}:7=\frac{3}{10}\)

\(3:\frac{3}{10}=\frac{1}{10}\)

từ đó ta có các tỉ lệ thức bằng nhau là:

28:14=8:4

3:10=2,1:7

1 tháng 11 2015

a. \(0,1\left(2\right)=0+\frac{12-1}{90}=\frac{11}{90};0,1\left(3\right)=0+\frac{13-1}{90}=\frac{12}{90}=\frac{2}{15}\)

=> 0,1(2)+0,1(3)=11/90 + 2/15 = 11/90 + 12/90 = 23/90

b. \(0,7\left(3\right)=0+\frac{73-7}{90}=\frac{66}{90}=\frac{11}{15};0,3\left(7\right)=0+\frac{37-3}{90}=\frac{34}{90}=\frac{17}{45}\)

=> 0,7(3)+0,3(7)=11/15 + 17/45 = 33/45 + 17/45 = 50/45 = 10/9

1 tháng 11 2015

a)\(0,1\left(2\right)+0,1\left(3\right)\)

=\(0,1+0,0\left(2\right)+0,1+0,0\left(3\right)\)

=\(0,2+0,0\left(1\right).2+0,0\left(1\right).3\)

=\(0,2+0,0\left(1\right).5\)

=\(0,2+\frac{1}{90}.5\)

=\(\frac{3,6}{18}+\frac{1}{18}\)

=\(\frac{4,6}{18}\)

20 tháng 9 2018

1.  -1/30

2. -4/5

3. 13/10

4.  6 

Cái này bấm máy tính cũng đc nha bn 

20 tháng 9 2018

Cảm ơn bạn

9 tháng 1 2022

Answer:

\(\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\) (Mình đã sửa đề nhé.)

\(=\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)

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

\(=\frac{3}{7}\)

\(\frac{3}{8}.27\frac{1}{5}-51\frac{1}{5}.38+19\)

\(=\frac{3}{8}\left(-24\right).38+19\)

\(=-342+19\)

\(=-323\)

\(-\frac{2}{7}.\frac{21}{8}.-\frac{7}{2}\)

\(=\left(\frac{-2}{7}.\frac{-7}{2}\right)\frac{21}{8}\)

\(=\frac{21}{8}\)

\(\frac{3}{5}.\left(\frac{5}{3}-\frac{2}{7}\right)-\left(\frac{7}{3}-\frac{3}{7}\right).\frac{3}{5}\)

\(=\frac{3}{5}.\text{[}\left(\frac{5}{3}-\frac{2}{7}\right)-\left(\frac{7}{3}-\frac{3}{7}\right)\text{]}\)

\(=\frac{3}{5}.\text{[}\frac{5}{3}-\frac{2}{7}-\frac{7}{3}+\frac{3}{7}\text{]}\)

\(=\frac{3}{5}.\text{[}\left(\frac{5}{3}-\frac{7}{3}\right)-\left(\frac{2}{7}-\frac{3}{7}\right)\text{]}\)

\(=\frac{3}{5}.\text{[}\frac{-2}{3}-\frac{-1}{7}\text{]}\)

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

\(=\frac{3}{5}.\left(\frac{-14}{21}+\frac{3}{21}\right)\)

\(=\frac{3}{5}.\frac{-11}{21}\)

\(=\frac{3.\left(-11\right)}{5.21}\)

\(=\frac{-11}{5.7}=\frac{-11}{35}\)

Chúc bạn học tốt

16 tháng 7 2016

\(\frac{\left(0,6\right)^5.\left(0,3\right)^3}{\left(0,2\right)^6.\left(0,3\right)^7}\)

\(=\frac{\left(\frac{6}{10}\right)^5.\left(\frac{3}{10}\right)^3}{\left(\frac{2}{10}\right)^6.\left(\frac{3}{10}\right)^7}\)

\(=\frac{6^5.3^3.\frac{1}{10^8}}{2^6.3^7.\frac{1}{10^{13}}}\)

\(=\frac{2^5.3^5.3^3}{2^6.3^7.\frac{1}{10^5}}\)

\(=\frac{10^5.3}{2}\)

\(=150000\)

18 tháng 9 2021

Bài 1:

\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)

\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)

\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)

\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)

\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)

\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)

Bài 2: 

\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)

 

18 tháng 9 2021

thanks bạn nhayeu