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28 tháng 9 2016

a) \(A=1.2+2.3+3.4+...+29.30\)

\(\Rightarrow3A=1.2.3+2.3.\left(4-1\right)+3.4\left(5-2\right)+...+29.30\left(31-28\right)\)

\(\Rightarrow3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+29.30.31-28.29.30\)

\(\Rightarrow3A=29.30.31\)

\(\Rightarrow A=29.30.31:3\)

\(\Rightarrow A=29.10.31\)

\(\Rightarrow A=8990\)

 

28 tháng 9 2016

3A= 1.2.3+2.3.4+3.4.3 +......+ 29.30.3

3A= 1.2. ﴾3 ‐ 0﴿ + 2.3.﴾4 ‐ 1﴿ +3.4. ﴾5 ‐ 2﴿....... . 29.30. ﴾31 ‐ 28﴿

3A = ﴾1.2.3 + 2.3.4 + 3.4.5 +...... +18.20.21﴿ ‐ ﴾0.1.2 + 1.2.3 + 2.3.4 +.......+ 18.19.20﴿

3A = 29.30.31 ‐ 0.1.2

3A =26970‐0

3A= 26970

 A=26970:3

A = 8990.

Vậy A=8990

26 tháng 6 2020

help me

26 tháng 6 2020

a)  45/343

b)  -1/4

c)  24/13

d)  -4 

bạn cho mình đi 

31 tháng 3 2016

Từng bài thôi

10 tháng 4 2016

a ) \(\frac{5}{7}.\frac{-7}{9}-\frac{5}{7}.\frac{2}{9}-\frac{5}{9}\)

18 tháng 6 2017

a,  \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)

\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)

\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)

\(A=\frac{-7}{20}+\frac{1}{2}\)

\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)

18 tháng 6 2017

b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)

\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)

\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)

\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)

\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)

\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)

c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)

\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)

\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)

\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)

\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)

\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)

d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)

\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)

\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)

\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)

20 tháng 5 2018

a) Đặt \(A=\frac{7^{15}}{1+7+7^2+...+7^{14}}\)

Đặt \(B=1+7+7^2+...+7^{14}\)

\(\Rightarrow7B=7+7^2+...+7^{15}\)

\(\Rightarrow7B-B=6B=7^{15}-1\)

\(\Rightarrow B=\frac{7^{15}-1}{6}\)

\(\Rightarrow A=\frac{7^{15}-1+1}{\frac{7^{15}-1}{6}}=\left(7^{15}-1\right).\frac{6}{7^{15}-1}+\frac{6}{7^{15}-1}=6+\frac{6}{7^{15}-1}\)

Tự làm tiếp nha

21 tháng 5 2018

bạn giải nốt đi

12 tháng 6 2018

b, Ta có:\(\dfrac{1+3+3^2+.....+3^{10}}{1+3+3^2+.....+3^9}\) \(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3+3^2+...+3^{10}}{1+3+3^2+...+3^9}\)\(=\dfrac{1}{1+3+3^2+...+3^9}+\dfrac{3.\left(1+3+3^2+...+3^9\right)}{1+3+3^2+...+3^9}\)

\(=\dfrac{1}{1+3+3^2+...+3^9}+3< 4\)

\(\Rightarrow\) \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< 4\) \(\left(1\right)\)

Ta có :\(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5+5^2+...+5^{10}}{1+5+5^2+....+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+\dfrac{5.\left(1+5+5^2+...+5^9\right)}{1+5+5^2+...+5^9}\)

\(=\dfrac{1}{1+5+5^2+...+5^9}+5>5\)

\(\Rightarrow\) \(\dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}>5\) \(\left(2\right)\)

Từ \(\left(1\right)và\left(2\right)\)

\(\Rightarrow\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

Vậy \(\dfrac{1+3+3^2+...+3^{10}}{1+3+3^2+...+3^9}< \dfrac{1+5+5^2+...+5^{10}}{1+5+5^2+...+5^9}\)

12 tháng 6 2018

a, Đặt \(A\)\(=\dfrac{7^{15}}{1+7+7^2+...+7^{14}}\)

\(\Rightarrow\) \(\dfrac{1}{A}\) \(=\dfrac{1+7+7^2+...+7^{14}}{7^{15}}=\dfrac{1}{7^{15}}+\dfrac{7}{7^{15}}+\dfrac{7^2}{7^{15}}+...+\dfrac{7^{14}}{7^{15}}\)

\(=\dfrac{1}{7^{15}}+\dfrac{1}{7^{14}}+\dfrac{1}{7^{13}}+....+\dfrac{1}{7}\)

Đặt \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)

\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)

\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)

\(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)

\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)

Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)

8 tháng 5 2019

\(A=5+5^2+5^3+5^4+5^5+5^6+5^7+5^8+5^9\)

\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+\left(5^7+5^8+5^9\right)\)

\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+5^7\left(1+5+5^2\right)\)

\(=5.31+5^4.31+5^7.31=31.\left(5+5^4+5^7\right)\)chia hết cho 31

Vậy A chia 31 dư 0

8 tháng 5 2019

\(S=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+8}\)

\(=1+\frac{1}{\left(1+2\right).3.\frac{1}{2}}+\frac{1}{\left(1+3\right).3.\frac{1}{2}}+...+\frac{1}{\left(1+8\right).8.\frac{1}{2}}\)

\(=1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{8.9}\)

\(=1+2.\left(\frac{3-2}{2+3}+\frac{4-3}{3.4}+...+\frac{9-8}{8.9}\right)\)

\(=1+2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\right)\)

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

\(=1+2.\frac{7}{18}=1+\frac{7}{9}=\frac{16}{9}\)