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ML
9 tháng 5 2016
A=\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)+\(\frac{1}{132}\)
A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)
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
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 \(B=\dfrac{9^{15}}{1+9+9^2+...+9^{14}}\)
\(\Rightarrow\dfrac{1}{B}=\dfrac{1+9+9^2+...+9^{14}}{9^{15}}=\dfrac{1}{9^{15}}+\dfrac{9}{9^{15}}+\dfrac{9^2}{9^{15}}+...+\dfrac{9^{14}}{9^{15}}\)
\(=\dfrac{1}{9^{15}}+\dfrac{1}{9^{14}}+\dfrac{1}{9^{13}}+...+\dfrac{1}{9}\)
Mà \(\dfrac{1}{7^{15}}>\dfrac{1}{9^{15}};\dfrac{1}{7^{14}}>\dfrac{1}{9^{14}};\dfrac{1}{7^{13}}>\dfrac{1}{9^{13}};....;\dfrac{1}{7}>\dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{A}>\dfrac{1}{B}\) \(\Rightarrow A< B\)
Vậy\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)
2 tháng 4 2017
\(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\frac{-9}{2}+\frac{1}{7}.\frac{-9}{10}\)
\(=\frac{-9}{10}.\frac{5}{14}+\frac{-9}{10}.\frac{1}{2}+\frac{1}{7}.\frac{-9}{10}\)
\(=\frac{-9}{10}.\left(\frac{5}{14}+\frac{1}{2}+\frac{1}{7}\right)=\frac{-9}{10}.1=\frac{-9}{10}\)
AY
22 tháng 5 2017
\(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\frac{-9}{10}\)
\(=\frac{-9}{10}.\frac{5}{14}+\frac{1}{7}.\frac{-9}{10}+\frac{1}{10}.\left(\frac{-9}{2}\right)\)
\(=\frac{-9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(\frac{-9}{2}\right)\)
\(=\frac{-9}{10}.\frac{1}{2}+\frac{1}{10}.\left(\frac{-9}{2}\right)\)
\(=\left(\frac{-9}{10}+\frac{1}{2}\right)^2\)
\(=\left(\frac{-9}{10}+\frac{5}{10}\right)^2\)
\(=\left(\frac{-2}{5}\right)^2\)
\(=\frac{4}{25}\)
DT
Tính nhanh
\(\frac{-9}{10}\).\(\frac{5}{14}+\frac{1}{10}.\frac{-9}{2}\)+ \(\frac{1}{7}.\frac{-9}{2}\)
2
B
5 tháng 4 2018
\(=\frac{-9}{2}.\frac{5}{70}+\frac{1}{10}.\frac{-9}{2}+\frac{1}{7}.\frac{-9}{2}\Rightarrow\)\(\frac{-9}{2}.\left(\frac{5}{70}+\frac{1}{10}+\frac{1}{7}\right)\Rightarrow\frac{-9}{2}.\frac{11}{35}=\frac{-99}{70}\)
29 tháng 7 2019
a) \(A=\frac{-7}{813}+496.\left(\frac{-7}{813}\right)+\left(\frac{-7}{813}\right).316\)
\(=\frac{-7}{813}.\left(1+496+316\right)\)
\(=\frac{-7}{813}.813\)
\(=-7\)
b) \(B=\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(\frac{-9}{10}\right)\)
\(=\frac{-9}{10}.\left(\frac{5}{14}+\frac{1}{2}+\frac{1}{7}\right)\)
\(=\frac{-9}{10}.1\)
\(=\frac{-9}{10}\)
1 tháng 5 2017
B1: Tính nhanh:
\(E=\dfrac{-9}{10}\cdot\dfrac{5}{14}+\dfrac{1}{10}\cdot\dfrac{-9}{2}+\dfrac{1}{7}\cdot\dfrac{-9}{10}\)
\(E=\dfrac{-9}{10}\cdot\dfrac{5}{14}+\dfrac{-9}{10}\cdot\dfrac{1}{2}+\dfrac{1}{7}\cdot\dfrac{-9}{10}\)
\(E=\dfrac{-9}{10}\cdot\left(\dfrac{5}{14}+\dfrac{1}{2}+\dfrac{1}{7}\right)\)
\(E=\dfrac{-9}{10}\cdot\left(\dfrac{5}{14}+\dfrac{7}{14}+\dfrac{2}{14}\right)\)
\(E=\dfrac{-9}{10}\cdot1=\dfrac{-9}{10}\)
B2: Chứng tỏ rằng:
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}< 1\)
Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Leftrightarrow1-\dfrac{1}{100}=\dfrac{99}{100}\)
Mà \(\dfrac{99}{100}< 1\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}< 1\)
Tick mình nha!
Ta có: \(-\frac{9}{10}\cdot\frac{5}{14}+\frac12\cdot\left(-\frac{9}{10}\right)+\frac17\cdot\frac{-1}{10}\)
\(=-\frac{9}{10}\left(\frac{5}{14}+\frac12\right)+\frac{-1}{7}\cdot\frac{1}{10}\)
\(=-\frac{9}{10}\left(\frac{5}{14}+\frac{7}{14}\right)-\frac{1}{70}\)
\(=-\frac{9}{10}\cdot\frac{12}{14}-\frac{1}{70}=\frac{-9}{10}\cdot\frac67-\frac{1}{70}=\frac{-54-1}{70}=-\frac{55}{70}=-\frac{11}{14}\)
\(-\frac{9}{10}\times\frac{5}{14}+\frac12\times\left(-\frac{9}{10}\right)+\frac17\times\left(-\frac{1}{10}\right)\)
\(=-\frac{9}{10}\times\left(\frac{5}{14}+\frac12\right)+\frac17\times\left(-\frac{1}{10}\right)\)
=\(-\frac{9}{10}\times\frac67+\frac17\times\left(-\frac{1}{10}\right)\)
\(=-\frac{27}{35}+\left(-\frac{1}{70}\right)\)
\(=-\frac{11}{14}\)