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A = \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) + ...........+ \(\dfrac{2}{99.101}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) +............+ \(\dfrac{1}{99}\) - \(\dfrac{1}{101}\)
A = \(\dfrac{1}{1}\) - \(\dfrac{1}{101}\)
A = \(\dfrac{100}{101}\)
A=21.3+23.5+25.7+...+299.101=1−13+13−15+15−17+...+199−1101=1−1101=100101
A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101
A = 1/1 - 1/101
A = 101/101 - 1/101
A = 100/101
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
A=21.3+23.5+25.7+...+299.101=1−13+13−15+15−17+...+199−1101=1−1101=100101
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
3−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
3−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
3−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
3−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
=31.3−11.3+53.5−33.5+75.7−55.7+…+10199.101−9999.101=1.33−1.31+3.55−3.53+5.77−5.75+…+99.101101−99.10199
=1−13+13−15+15−17+…+199−1101=1−31+31−51+51−71+…+
Đúng(0)
1.33−1+3.55−3+5.77−5+…+99.101101−99
\(= \frac{3}{1.3} - \frac{1}{1.3} + \frac{5}{3.5} - \frac{3}{3.5} + \frac{7}{5.7} - \frac{5}{5.7} + \ldots + \frac{101}{99.101} - \frac{99}{99.101}\)
\(= 1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \ldots + \frac{1}{99} - \frac{1}{101}\)
\(= 1 - \frac{1}{101} = \frac{100}{101}\)
Vậy \(\frac{2}{1.3} + \frac{2}{3.5} + \frac{2}{5.7} + \ldots + \frac{2}{99.101} = \frac{100}{101}\)
K
Uurur
Ttgct
V
Đáp án : 100/101
2/.3 + 2/3.5 + 2/5.7 + ... + 2/99.101 =
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99
= 1 - 101
= 100/101
tính tổng A=\(\dfrac{2}{1.3}\)+\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+...+\(\dfrac{2}{99.101}\)
`A=2/[1.3]+2/[3.5]+2/[5.7]+.....+2/[99.101]`
`A=1-1/3+1/3-1/5+1/5-1/7+......+1/99-1/101`
`A=1-1/101=101-1/101=100/101`
\(\dfrac{100}{101}\)
B= \(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) +\(\dfrac{2}{5.7}\) +...+ \(\dfrac{2}{97.99}\) + \(\dfrac{2}{99.101}\)
\(B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{97\cdot99}+\dfrac{2}{99\cdot101}\\ B=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\\ B=\dfrac{1}{1}-\dfrac{1}{101}\\ B=\dfrac{101}{101}-\dfrac{1}{101}\\ B=\dfrac{100}{101}\)
\(\dfrac{2}{1\cdot3}=\dfrac{1}{1}-\dfrac{1}{3}=\dfrac{3}{3}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\dfrac{2}{3\cdot5}=\dfrac{1}{3}-\dfrac{1}{5}=\dfrac{5}{15}-\dfrac{3}{15}=\dfrac{2}{15}\)
\(\dfrac{2}{5\cdot7}=\dfrac{1}{5}-\dfrac{1}{7}=\dfrac{7}{35}-\dfrac{5}{35}=\dfrac{2}{35}\)
và cứ như thế đến số cuối
Tính tổng
a) \(\dfrac{2}{1.3}\)+ \(\dfrac{2}{3.5}\)+ \(\dfrac{2}{5.7}\)+......+\(\dfrac{2}{99.101}\)
b) \(\dfrac{2}{1.3}\)+ \(\dfrac{5}{3.5}\)+ \(\dfrac{5}{5.7}\)+.......+ \(\dfrac{5}{99.101}\)
Trả lời
a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...\dfrac{2}{99.101}\)
=\(2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)
=\(2.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(2.\left(\dfrac{1}{1}-\dfrac{1}{101}\right)\)
=\(2.\dfrac{100}{101}\)
=\(\dfrac{200}{101}\)
Hình như phần b bạn chép đề sai hay sao đấy
Tính nhanh:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
B = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+....+\dfrac{2}{99.101}\)
\(M=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{3}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{3}{4}\) \(\frac{3}{4}\) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=2-\frac{2}{101}=\frac{200}{101}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(B=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(B=2.\frac{100}{101}=\frac{200}{101}\)
tính tổng: A= \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\) B= \(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{3.7}+...+\dfrac{5}{99.101}\)
C= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\) D= \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\) E= \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
Tinh nhanh:
\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\) + \(\dfrac{2}{5.7}\) +...+ \(\dfrac{2}{99.101}\) b) \(\dfrac{5}{1.3}\) +\(\dfrac{5}{3.5}\) + \(\dfrac{5}{5.7}\) + ... + \(\dfrac{5}{99.101}\)
HELP ME
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}\\ =\dfrac{100}{101}\)
\(\dfrac{5}{1\cdot3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{99\cdot101}\\ =\dfrac{5}{2}\cdot\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\\ =\dfrac{5}{2}\cdot\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\left(1-\dfrac{1}{101}\right)\\ =\dfrac{5}{2}\cdot\dfrac{100}{101}\\ =\dfrac{250}{101}\)
\(a,\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...\dfrac{1}{99}-\dfrac{1}{101}\)
\(=1-\dfrac{1}{101}\)
\(=\dfrac{100}{101}\)
\(\dfrac{2}{3.5}+\dfrac{2}{5.7}\dfrac{2}{7.9}+.........+\dfrac{2}{99.101}\)
\(P=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}\)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
tính hợp lý:
\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\\ =\dfrac{4}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\\ =2.\left(1-\dfrac{1}{101}\right)\\ =2.\dfrac{100}{101}\\ =\dfrac{200}{101}\)
`4/1.3+4/3.5+4/5.7+...+4/99.101`
`=2(2/1.3+2/3.5+2/5.7+...+2/99.101)`
`=2(1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101)`
`=2(1-1/101)`
`=2. 100/101`
`=200/101`
1/.\(\dfrac{1}{1}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}\)
2/.\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{10100}\)
3/.A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
4/.A = \(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)
tính bằng cách thuận tiện nhất ( làm nhanh trước 5h nha , nếu ai làm được thì cho 100 tick , thật đó và trình bày cách diễn giải nha )
2/ = \(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) +......+\(\dfrac{1}{100.101}\)
= 1-\(\dfrac{1}{2}\) +\(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)+.........+\(\dfrac{1}{100}\)-\(\dfrac{1}{101}\)
=1-\(\dfrac{1}{101}\)=...........
mk làm vậy thôi nha
thông cảm
=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{4.5}\)=\(1-\dfrac{1}{2}+....+\dfrac{1}{4}-\dfrac{1}{5}\)
=1-\(\dfrac{1}{5}=\dfrac{4}{5}\)
tương tự
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