M=\(\frac{1}{15}\)+\(\frac{1}{105}\)+\(\frac{1}{315}\)+...+\(\frac{1}{9177}\)
So sánh M với \(\frac{1}{12}\)
Những câu hỏi liên quan
cho M = \(\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+....+\frac{1}{9177}\). so sánh M với \(\frac{1}{12}\)
\(M=\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+...+\frac{1}{9177}\)
\(M=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{19.21.23}\)
\(M=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{19.21}-\frac{1}{21.23}\right)\)
\(M=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{21.23}\right)< \frac{1}{4}.\frac{1}{1.3}=\frac{1}{12}\)
\(\Rightarrow M< \frac{1}{12}\)
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1) Cho M=\(\frac{1}{15}\)+\(\frac{1}{105}\)+\(\frac{1}{315}\)+...+\(\frac{1}{9177}\)
So sánh M với \(\frac{1}{12}\)
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! )
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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
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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 Mfrac{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)Afrac{3}{x-1}...
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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
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: \(N=\dfrac{1095}{547}\cdot\dfrac{3}{211}-\dfrac{546}{547\cdot211}-\dfrac{4}{547\cdot211}\)
\(=\dfrac{1}{547\cdot211}\left(1095\cdot3-546-4\right)\)
\(=\dfrac{1}{547\cdot211}\cdot2735=\dfrac{5}{211}\)
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Cho M= 1/5 +1/105 +1/315+...+1/9177. So sanh M vs 12
2frac{1}{315}.frac{1}{651}-frac{1}{105}.3frac{650}{651}-frac{4}{315.651}+frac{4}{105}left(2+frac{1}{315}right).frac{1}{651}-frac{1}{105}.left(3+frac{650}{651}right)-frac{4}{315.651}+frac{4}{105}2.frac{1}{651}+frac{1}{315}.frac{1}{651}-frac{1}{105}.3+frac{1}{105}.frac{650}{651}-frac{4}{315.651}+frac{4}{105}frac{2}{651}+frac{1}{315.651}-frac{3}{105}-frac{650}{105.651}-frac{4}{315.651}+frac{4}{105}frac{2}{615}+left(frac{1}{315.651}-frac{4}{315.651}right)+left(frac{-3}{105}+frac{4}{105}right)-frac{6...
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\(2\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\left(2+\frac{1}{315}\right).\frac{1}{651}-\frac{1}{105}.\left(3+\frac{650}{651}\right)-\frac{4}{315.651}+\frac{4}{105}\)
\(=2.\frac{1}{651}+\frac{1}{315}.\frac{1}{651}-\frac{1}{105}.3+\frac{1}{105}.\frac{650}{651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{651}+\frac{1}{315.651}-\frac{3}{105}-\frac{650}{105.651}-\frac{4}{315.651}+\frac{4}{105}\)
\(=\frac{2}{615}+\left(\frac{1}{315.651}-\frac{4}{315.651}\right)+\left(\frac{-3}{105}+\frac{4}{105}\right)-\frac{650}{105.651}\)
\(=\frac{2}{651}-\frac{3}{315.651}+\frac{1}{105}-\frac{650}{105.651}\)
\(=\left(\frac{2}{651}+\frac{1}{105}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\left(\frac{2.105}{105.651}+\frac{651}{105.651}\right)-\frac{650}{105.651}-\frac{3}{315.651}\)
\(=\frac{211}{105.651}-\frac{3}{315.651}\)
\(=\frac{1}{651}.\left(\frac{211}{105}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.\left(\frac{633}{315}-\frac{3}{315}\right)\)
\(=\frac{1}{651}.2\)
\(=\frac{2}{651}\)