(1/3)^2 . 1/3 . 9^2
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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
21 tháng 4 2020
B)2-9+1-3
.vì bỏ ngoặc trước nó là dấu trừ thì ta đổi dấu các số hạng trong ngoặc
27 tháng 7 2021
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
NH
0
NH
2
16 tháng 11 2021
ta có: \(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^9}=1\)
mà \(1+3+3^2+...+3^9>1+3+3^2+...+3^8\)
\(\Rightarrow B=\frac{1+3+3^2+...+3^9}{1+3+3^2+...+3^8}>1\)
\(\Rightarrow A< B\)
16 tháng 11 2021
Ta thấy : A= ( 1+5+5^2+.......+5^9)/(1+5+5^2+...... +5^8)= 5^9
B=(1+3+3^2+......+3^9)/(1+3+3^2+,,,,,,,,+3/9)=1
mÀ 5^9 > 1 . SUY RA A>B
Vậy A>B
mk ko chắc chắn lắm
k cho mk nhé
\(\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3}\cdot9^2\\ =\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3}\cdot\left(3^2\right)^2\\ =\left(\dfrac{1}{3}\right)^{1+2}\cdot3^{2\cdot2}\\ =\left(\dfrac{1}{3}\right)^3\cdot3^4\\ =\dfrac{1}{3^3}\cdot3^4\\ =\dfrac{3^4}{3^3}\\ =3\)
\(\left(\dfrac{1}{3}\right)^2.\dfrac{1}{3}.9^2=\dfrac{1}{9}.\dfrac{1}{3}.81=\dfrac{81}{27}=\dfrac{9}{3}=3\)