Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(A=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(A=\frac{84}{305}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Bài làm
\(S=\frac{7}{3.5}+\frac{7}{5.7}+...+\frac{7}{59.61}\)
\(S=7\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(S=7\left(\frac{1}{3}-\frac{1}{61}\right)\)
\(S=7\left(\frac{61}{183}-\frac{3}{183}\right)\)
\(S=7.\frac{58}{183}\)
\(S=\frac{406}{183}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Dấu chấm là nhân
a) \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{99.100}\) \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
b) \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\) \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}=1-\frac{1}{99}=\frac{98}{99}\)
c) Đặt \(C=\frac{4}{5.7}+\frac{4}{7.9}+....+\frac{4}{59.61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)
\(\Rightarrow C=\frac{56}{305}:\frac{1}{2}=\frac{112}{305}\)
CHÚC BẠN HỌC TỐT NHA! ĐÚNG THÌ NHA!
B = 1/4 x( 4/ 1x3x5 +4 /3x5x7 +...+ 4/47x49x51)
= 1/4 x( 1/ 1x3 - 1/3x5 +1/3x5 -1/5x7 +...+ 1/47x49 -1/49x51)
= 1/4 x( 1/1x3 -1/49x51)
= 1/4 x( 1/3 - 1/2499)
= 1/4 x 832/2499
= 208 /2499
Vậy B= 208 /2499
C= 1/ 9x11 -1/ 11x13 +1/ 11x13 -1/ 13x15 +... + 1/ 59x61 -1/61x63
= 1/ 9x11 -1/ 61x63
= 1/99 -1/3843
= 416 /42273
Vậy C= 416 /42273
B = \(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+....+\frac{3}{59\cdot61}\)
=> B = \(\frac{1}{2}\cdot\left(\frac{3}{5}-\frac{3}{7}\right)+\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{3}{9}\right)+...+\frac{1}{2}\cdot\left(\frac{3}{59}-\frac{3}{61}\right)\)
=>B= \(\frac{1}{2}\cdot\left(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+....+\frac{3}{59}-\frac{3}{61}\right)\)
=> B = \(\frac{1}{2}\cdot\left(\frac{3}{5}-\frac{3}{61}\right)\)= \(\frac{1}{2}\cdot\frac{168}{305}\)= \(\frac{84}{305}\) Vậy B= \(\frac{84}{305}\)
\(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)(đã sửa)
Đặt \(A=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{2}{3}\left(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+....+\dfrac{3}{59\cdot61}\right)\)
\(\dfrac{2}{3}A=\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\)
\(\dfrac{2}{3}A=\dfrac{1}{5}-\dfrac{1}{61}\)
\(A=\dfrac{56}{305}:\dfrac{2}{3}=\dfrac{56}{305}\cdot\dfrac{3}{2}=\dfrac{84}{305}\)
S=4/5.7+4/7.9+..+4/59.61
=>S=4/2.(2/5.7+2/7.9+...+2/59.61)
=>S=2.(1/5-1/7+1/7-1/9+...+1/59-1/61)
=>S=2.(1/5-1/61)=2.56/305=112/305
vậy S=112/305
