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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\(\dfrac{7^{15}}{1+7+7^2+...+7^{14}}>\dfrac{9^{15}}{1+9+9^2+....+9^{14}}\)
26 tháng 3 2017
b, B = 1 + 2 + 2^2 + 2^3 +.....+ 2^2013
2B = 2.(1 + 2 + 2^2 + 2^3 +.....+ 2^2013)
2B = 2 + 2^2 + 2^3 + 2^4 +.....+ 2^2014
2B - B = 2^2014 - 1
B = 2^2014 - 1
27 tháng 7 2019
\(1+3+5+...+\left(2n-1\right)\)
\(=\frac{\left[\left(2n-1-1\right):2+1\right]\left(2n-1+1\right)}{2}\)
\(=\frac{\left[\left(2n-2\right):2+1\right]2n}{2}\)
\(=\frac{\left(n-1+1\right)2n}{2}\)
\(=\frac{n.2n}{2}\)
\(=\frac{2n^2}{2}\)
\(=n^2\)
TT
0
KK
11 tháng 9 2018
4^3^10=4^30=(4^2)^15=..........6^15=...........6
2^2^5=2^10=(2^4)^2 . 2^2=...........6^2 . ...........4=.............4
2^3^4=2^12=(2^4)^3=.............6^3=...............6
3^3^3=3^9=(3^4)^2 . 3=..............1^2 . 3=..............3
9^9^9=9^81=(9^2)^80 . 9=..............1^80 . 9=.................9
24 tháng 3 2019
a, \(\frac{9}{1.2}+\frac{9}{2.3}+...+\frac{9}{99.100}\)
=9.(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\))
= 9(1 -\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\))
=9(1-\(\frac{1}{100}\))
A=\(\frac{891}{100}\)
b, \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{27.30}\)
=1-(\(\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{27}-\frac{1}{30}\))
=1-\(\frac{1}{30}\)
B=\(\frac{29}{30}\)
24 tháng 3 2019
a) \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+...+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)
b) \(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{27.30}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{27}-\dfrac{1}{30}\)
\(=1-\dfrac{1}{30}\)
\(=\dfrac{29}{30}\)
9/2+2/3
=27/6+4/6
=31/6
\(\frac92+\frac23\)
\(=\frac{27}{6}+\frac46\)
\(\frac{27+4}{6}=\frac{31}{6}\)