1) Cho M=\(\frac{1}{15}\)+\(\frac{1}{105}\)+\(\frac{1}{315}\)+...+\(\frac{1}{9177}\)
So sánh M với \(\frac{1}{12}\)
cho M = \(\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+....+\frac{1}{9177}\). so sánh M với \(\frac{1}{12}\)
\(M=\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+...+\frac{1}{9177}\)
\(M=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{19.21.23}\)
\(M=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{19.21}-\frac{1}{21.23}\right)\)
\(M=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{21.23}\right)< \frac{1}{4}.\frac{1}{1.3}=\frac{1}{12}\)
\(\Rightarrow M< \frac{1}{12}\)
M=\(\frac{1}{15}\)+\(\frac{1}{105}\)+\(\frac{1}{315}\)+...+\(\frac{1}{9177}\)
So sánh M với \(\frac{1}{12}\)
\(\text{M}< \frac{1}{12}\)
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Cho M=1/15+1/105+1/315+...+1/9177
So sánh M với 1/12
Cho M = 1/15+1/105+1/315+.............+1/9177. So sánh M với 1/2
Cho M= 1/15 + 1/105 + 1/315+....+1/9177. So sanh M vs 12.
Help me!
S=1/1.3.5 +1/3.5.7+...................+1/19.21.23
---> 4S=4/1.3.5 + 4/ 3.5.7 +.........................+4/19.21.23
= (1/1.3 -1/3.5 ) + ( 1/3.5 -1/3.7) +...................................+(1/... -1/21.23 )
= 1/1.3 -1/21.23
= 1/3 -1/483 =160/483
----> S = 160/483 : 4 = 160 / 1932
(em tự rút gọn nhé! )S=1/1.3.5 +1/3.5.7+...................+1/19.21.23
---> 4S=4/1.3.5 + 4/ 3.5.7 +.........................+4/19.21.23
= (1/1.3 -1/3.5 ) + ( 1/3.5 -1/3.7) +...................................+(1/... -1/21.23 )
= 1/1.3 -1/21.23
= 1/3 -1/483 =160/483
----> S = 160/483 : 4 = 160 / 1932
(em tự rút gọn nhé! )
M=1/1.3.5 +1/3.5.7+.....+1/19.21.23
4M=4/1.3.5 + 4/ 3.5.7 +...+4/19.21.23
= (1/1.3 -1/3.5 ) + ( 1/3.5 -1/3.7) +....+(1/19.21 -1/21.23 )
= 1/1.3 -1/21.23
= 1/3 -1/483
=160/483
⇒ S = 160/483 : 4
= 160 / 1932
=40/438
Bài 1:Chứng tỏ rằng
a)\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2009.2010}< 1\)
b)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 1\)
c)\(\frac{2}{5}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}< \frac{8}{9}\)
d)\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Bài 2:Cho M=\(\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+..+\frac{1}{9177}\).So sánh với 12
Bài 3:Với giá trị nào của x \(\in\) Z các phân số sau có giá trị là 1 số nguyên
a)A=\(\frac{3}{x-1}\) b)B=\(\frac{x-2}{x+3}\) c)C=\(\frac{2x+1}{x-3}\) d)D=\(\frac{x^2-1}{x+1}\)
Bài 4:a) Chứng tỏ rằng các phân số sau tối giản với mọi số tự nhiên n
a)\(\frac{n+1}{2n+3}\) b)\(\frac{2n+3}{4n+8}\)
Mình đang cần gấp lắm ,làm ơn
Cho M= 1/5 +1/105 +1/315+...+1/9177. So sanh M vs 12
Tính giá trị của các biểu thức sau
a) M= \(2\frac{1}{315}\cdot\frac{1}{651}-\frac{1}{105}\cdot3\frac{650}{651}-\frac{4}{315\cdot651}+\frac{4}{105}\)
b) N= \(2\frac{1}{547}\cdot\frac{3}{211}-\frac{546}{547}\cdot\frac{1}{211}-\frac{4}{547\cdot211}\)
a: \(M=\dfrac{631}{315}\cdot\dfrac{1}{651}-\dfrac{1}{105}\cdot\dfrac{2603}{651}-\dfrac{4}{315\cdot651}+\dfrac{4}{105}\)
\(=\dfrac{1}{315\cdot651}\cdot\left(631-4\right)-\dfrac{1}{105}\left(\dfrac{2603}{651}-4\right)\)
\(=\dfrac{1}{105}\cdot\dfrac{1}{1953}\cdot627+\dfrac{1}{105\cdot651}\)
\(=\dfrac{1}{105\cdot651}\left(\dfrac{1}{3}\cdot627+1\right)=\dfrac{1}{105\cdot651}\cdot210=\dfrac{2}{651}\)
b: \(N=\dfrac{1095}{547}\cdot\dfrac{3}{211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{1}{547\cdot211}\left(1095\cdot3-546-4\right)\)
\(=\dfrac{1}{547\cdot211}\cdot2735=\dfrac{5}{211}\)
a, Cho A=\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{99}+\frac{1}{100}\) . So Sánh A với 1
b, B=\(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\). So sánh B với \(\frac{1}{2}\)
c, cho M=\(\frac{2013}{2014}+\frac{2014}{2015}\)và N=\(\frac{2013+2014}{2014+2015}\). So sánh M và N
Câu a, p/s cuối cùng là \(\frac{1}{100}\)nha mí bn
a) Ta có :
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}\)
\(>\frac{1}{10}+\frac{1}{100}.90=\frac{1}{10}+\frac{90}{100}=1\)
vậy A > 1
b) \(B=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\)
\(>\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{20}.10=\frac{1}{2}\)
Vậy B > \(\frac{1}{2}\)