(9/5) ⁷ ÷ (9/5) ² = (9/5) ²a-1
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VN
Võ Ngọc Phương
VIP
25 tháng 4 2023
Ta có:
\(\dfrac{9}{5}>1\) ; \(\dfrac{9}{9}=1\)
\(\dfrac{5}{9}>1\) ; \(\dfrac{1}{3}>1\)
⇔ \(\dfrac{1}{3}=\dfrac{3}{9}\)
mà \(\dfrac{5}{9}>\dfrac{3}{9}\)
⇒ Chọn câu C
Chúc bạn học tốt
11 tháng 8 2016
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 \(\frac{25}{11}\)
11 tháng 8 2016
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
KM
3
QL
1
20 tháng 8 2023
\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}-\dfrac{9}{11}-\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\left(+\dfrac{7}{9}\rightarrow-\dfrac{7}{9}\right)\)
\(\Rightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}+\dfrac{7}{9}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}\)
\(\Rightarrow A=-\dfrac{11}{13}+\dfrac{13}{15}\)
\(\Rightarrow A=\dfrac{-11.15+13.13}{13.15}\)
\(\Rightarrow A=\dfrac{-165+169}{195}=\dfrac{4}{195}\)
6 tháng 4 2018
2/9+3/9+4/9+5/9+6/9+7/9
=2+3+4+5+6+7/9
=27/9
=3
1/6+4/9+5/6+11/16+5/9+5/16
=1/6+5/6+4/9+5/9+11/16+5/16
=1+1+1=3
hok tốt ~
29 tháng 4 2017
Ta có :
+) \(A=\dfrac{1+9+9^2+...+9^{2009}}{1+9+9^2+...+9^{2009}}+\dfrac{9^{2010}}{1+9+9^2+...+9^{2009}}\)
\(A=1+1:\dfrac{1+9+9^2+...+9^{2009}}{9^{2010}}\)
\(A=1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)\)
+) \(B=\dfrac{1+5+5^2+...+5^{2009}}{1+5+5^2+...+5^{2009}}+\dfrac{5^{2010}}{1+5+5^2+...+5^{2009}}\)
\(B=1+1:\dfrac{1+5+5^2+...+5^{2009}}{5^{2010}}\)
\(B=1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
Vì \(\dfrac{1}{9^{2010}}< \dfrac{1}{5^{2010}}\)
\(\dfrac{1}{9^{2009}}< \dfrac{1}{5^{2009}}\) (ngoặc cả mấy cài so sánh này vào rôi mời suy ra nhé)
.............................
\(\dfrac{1}{9}< \dfrac{1}{5}\)
\(\)=> \(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}< \dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\)
=> \(1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
=> \(1+1:\left(\dfrac{1}{9^{2010}}+\dfrac{1}{9^{2009}}+...+\dfrac{1}{9}\right)>1+1:\left(\dfrac{1}{5^{2010}}+\dfrac{1}{5^{2009}}+...+\dfrac{1}{5}\right)\)
Hay A > B
16 tháng 1 2022
Bài 1:1×2×3×4×5×6×7×8×9×10 bằng mấy? Bài 2:5×5×5×5×5×5×5×5×5×5=3628800
Bài 2:9×9×9×9×9×9×9×9×9×9 = 3486784401 (bạn k cho mình nha)
D
10 tháng 5 2015
A = \(1+\frac{9^{2010}}{1+9+9^2+....+9^{2009}}\)= \(1+1:\frac{1+9+9^2+....+9^{2009}}{9^{2010}}\)= \(1+1:\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+\frac{1}{9^{2008}}+...+\frac{1}{9}\right)\)
B = \(1+\frac{5^{2010}}{1+5+5^2+....+5^{2009}}\)= \(1+1:\frac{1+5+5^2+...+5^{2009}}{5^{2010}}\)= \(1+1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)
Do \(\frac{1}{9^{2010}}<\frac{1}{5^{2010}}\) ; \(\frac{1}{9^{2009}}<\frac{1}{5^{2009}}\) ;.....; \(\frac{1}{9}<\frac{1}{5}\)
=> \(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+...+\frac{1}{9}<\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\)
=> 1:\(\left(\frac{1}{9^{2010}}+\frac{1}{9^{2009}}+...+\frac{1}{9}\right)>1:\left(\frac{1}{5^{2010}}+\frac{1}{5^{2009}}+...+\frac{1}{5}\right)\)
Vậy A > B
\(\left(\dfrac{9}{5}\right)^7:\left(\dfrac{9}{5}\right)^2=\left(\dfrac{9}{5}\right)^2a-1\)
\(\Leftrightarrow a\cdot\dfrac{81}{25}-1=\left(\dfrac{9}{5}\right)^5\)
\(\Leftrightarrow a\cdot\dfrac{81}{25}=\dfrac{59049}{3125}+1=\dfrac{62174}{3125}\)
\(\Leftrightarrow a=\dfrac{686}{1125}\)
(95)7:(95)2=(95)2a−1(95)7:(95)2=(95)2a−1
=a⋅8125−1=(95)5⇔a⋅8125−1=(95)5
=a⋅8125=590493125+1=621743125=a⋅8125=590493125+1=621743125
=a=6861125