B=1^2+3^2+5^2+7^2+...+101^2
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2 tháng 3 2017
\(\frac{2}{1.2}+\frac{2}{3.5}+\frac{2}{5.7}+......+\frac{2}{99.101}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
NB
1
Q
1 tháng 4 2017
\(B=\dfrac{2^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{2^2}{5.7}+...+\dfrac{2^2}{99.101}\)
\(B=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(B=\dfrac{1}{2}\left(1-\dfrac{1}{101}\right)\)
\(B=\dfrac{1}{2}.\dfrac{100}{101}\)
\(B=\dfrac{50}{101}\)
30 tháng 3 2015
A = 2/1*5 + 2/5*9 + ... + 2/101*105
= 1/2(4/1*5 + 4/5*9 + ... + 4/101*105)
= 1/2(1 - 1/5 + 1/5 - 1/9 + ... + 1/101 - 1/105)
= 1/2(1 - 1/105)
= 1/2 * 104/105 = 52/105
Sửa câu b. Phân số thứ 2 phải là 4/5*8
B = 4/2*5 + 4/5*8 + ... + 4/47*50
= 4/3(3/2*5 + 3/5*8 + ... + 3/47*50)
= 4/3(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/47 - 1/50)
= 4/3(1/2 - 1/50)
= 4/3 * 24/50 = 16/25
17 tháng 4 2019
\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
\(B=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(B=\frac{5}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(B=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(B=\frac{5}{2}.\left(1-\frac{1}{101}\right)\)
\(B=\frac{5}{2}.\frac{100}{101}\)
\(B=\frac{250}{101}\)
31 tháng 12 2017
1
b;
B=1+ (7-5) + (11-9) + ...+(101-99)
B=1+2+2+..+2
B=1+25.2=51
NT
31 tháng 12 2017
2.
a.
ĐK : x+2 >=0 => x>=-2
\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)
Vậy x=-2
LT
3
13 tháng 5 2016
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+....+\frac{2}{99\cdot101}\)
\(\frac{2}{1\cdot3}=\frac{3-1}{1\cdot3}=\frac{3}{1\cdot3}-\frac{1}{1\cdot3}=\frac{1}{1}-\frac{1}{3}=1-\frac{1}{3}\)
\(\frac{2}{3\cdot5}=\frac{5-3}{3\cdot5}=\frac{5}{3\cdot5}-\frac{3}{3\cdot5}=\frac{1}{3}-\frac{1}{5}\)
....
\(\frac{2}{99\cdot101}=\frac{101-99}{99\cdot101}=\frac{101}{99\cdot101}-\frac{99}{99\cdot101}=\frac{1}{99}-\frac{1}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)
13 tháng 5 2016
\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+\frac{5}{5\cdot7}+...+\frac{5}{99\cdot101}\)
=\(\frac{5}{2}\cdot\frac{2}{1\cdot3}+\frac{5}{2}\cdot\frac{2}{3\cdot5}+\frac{5}{2}\cdot\frac{2}{5\cdot7}+...+\frac{5}{2}\cdot\frac{2}{99\cdot101}\)
=\(\frac{5}{2}\cdot\left[\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right]\)
=\(\frac{5}{2}\cdot\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right]\)
=\(\frac{5}{2}\cdot\left(1-\frac{1}{101}\right)\)
=\(\frac{5}{2}\cdot\frac{100}{101}\)
\(=\frac{250}{101}\)
13 tháng 8 2023
a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
TD
16 tháng 2 2021
B=1+2-(3+4)+5+6-..-100+101
B=(3+11+19+...+195)-(7+15+...+199)+101
B=25.99-25.103+101
B=-100+101=1
Vậy B=1
\(1^2+3^2+5^2+\cdots+101^2\)
Số hạng có trong dãy trên là:
\(\frac{101-1}{2}+1\) = 51 (số hạng)
\(\frac{51\cdot\left(2\cdot51-1\right)\cdot\left(2+51-1\right)}{3}\)
\(=\frac{51\cdot101\cdot103}{3}\)
\(=\frac{530553}{3}=176851\)