\(\dfrac{31}{2}\cdot\dfrac{32}{2}...\dfrac{60}{2}=1.3.5...59\)
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NT
1
26 tháng 5 2015
\(1.3.5.7.9...59=\frac{\left(1.3.5...59\right).\left(2.4.6...60\right)}{2.4.6...60}=\frac{1.2.3...60}{2^{30}\left(1.2.3...30\right)}\)
\(=\frac{31.32.33...60}{2.2.2...2}=\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}...\frac{60}{2}\)
Vậy \(\frac{31}{2}\cdot\frac{32}{2}\cdot\frac{33}{2}...\frac{60}{2}=1.3.5...59\)(đpcm)
25 tháng 3 2017
Giải:
Đặt \(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)
Ta có:
\(A=\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}\)
\(\Rightarrow A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}\right)+\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}\right)+\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}\right)\)
Nhận xét:
\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}< \dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}=\dfrac{1}{3}\)
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}< \dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)
\(\Rightarrow A< \dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{47}{60}< \dfrac{48}{60}=\dfrac{4}{5}\)
\(\Rightarrow A< \dfrac{4}{5}\left(1\right)\)
Lại có:
\(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{40}>\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}=\dfrac{1}{4}\)
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{50}>\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{1}{5}\)
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}=\dfrac{1}{6}\)
\(\Rightarrow A>\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac{3}{5}\)
\(\Rightarrow A>\dfrac{3}{5}\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{3}{5}< A< \dfrac{4}{5}\)
Vậy \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\) (Đpcm)
LT
22 tháng 4 2018
Đặt A=131+132+133+...+159+160A=131+132+133+...+159+160
Ta có:
A=131+132+133+...+159+160A=131+132+133+...+159+160
⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)⇒A=(131+132+...+140)+(141+142+...+150)+(151+152+...+160)
Nhận xét:
131+132+...+140<130+130+...+130=13131+132+...+140<130+130+...+130=13
141+142+...+150<140+140+...+140=14141+142+...+150<140+140+...+140=14
151+152+...+160<150+150+...+150=15151+152+...+160<150+150+...+150=15
⇒A<13+14+15=4760<4860=45⇒A<13+14+15=4760<4860=45
⇒A<45(1)⇒A<45(1)
Lại có:
131+132+...+140>140+140+...+140=14131+132+...+140>140+140+...+140=14
141+142+...+150>150+150+...+150=15141+142+...+150>150+150+...+150=15
151+152+...+160>160+160+...+160=16151+152+...+160>160+160+...+160=16
⇒A>14+15+16=3760>3660=35⇒A>14+15+16=3760>3660=35
⇒A>35(2)⇒A>35(2)
Từ (1)(1) và (2)(2)
⇒35<A<45⇒35<A<45
Vậy 35<131+132+133+...+159+160<4535<131+132+133+...+159+160<45
LM
13 tháng 6 2018
Ta có:\(\dfrac{31}{2}\).\(\dfrac{32}{2}\).\(\dfrac{33}{2}\).....\(\dfrac{60}{2}\)
=\(\dfrac{31.32.33.....60}{2^{30}}\)
=\(\dfrac{\left(1.2.3.....30\right).\left(31.32.33.....60\right)}{\left(1.2.3.....30\right).2^{30}}\)
=\(\dfrac{1.2.3.....60}{2.4.6.....60}\)
=\(\dfrac{\left(1.3.5.....59\right).\left(2.4.6.....60\right)}{2.4.6.....60}\)
=1.3.5.....59
Vậy (đpcm)
NA
Nguyễn An Ninh
VIP
22 tháng 4 2023
=\(\dfrac{11}{31}\)*\(\left(\dfrac{-2}{17}-\dfrac{-9}{17}\right)\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)
=\(\dfrac{11}{31}\)*\(\dfrac{-7}{17}\)+\(\dfrac{7}{31}\)*\(\dfrac{-20}{17}\)
=\(\dfrac{-77}{512}\)+\(\dfrac{-140}{512}\)
=\(\dfrac{-217}{512}\)
28 tháng 4 2017
Ta có :
\(\dfrac{31}{2}.\dfrac{32}{2}.\dfrac{33}{2}.....\dfrac{60}{2}=31.32.33.....\dfrac{60}{2^{30}}\)
(31.32.33....60)(1.2.3....30)/230(1.2.3....30)
= (1.3.5.....59)(2.4.6.....60 )/( 2.4.6....60 ) = 1.3.5....59
\(\Rightarrow P=Q\)
ko giống như cách của mk nhưng cx 1like
HP
10 tháng 1 2016
31/2.32/2.33/2....60/2=(31.32.33...60)/230
=[(31.32.33...60).(1.2.3...30)]/230.(1.2.3...30)
=[(1.3.5...59).(2.4.6....60)]/(2.4.6...60)=1.3.5...59
NP
0
