So sánh P và Q:
P= \(\dfrac{31}{2}.\dfrac{32}{2}.\dfrac{33}{2}.....\dfrac{60}{2}\)
và Q= \(1.3.5.7.....59\)
Giải thích rõ ràng nhé!
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Chứng tỏ rằng : \(\dfrac{31}{2}\) . \(\dfrac{32}{2}\) . \(\dfrac{33}{2}\) . .... . \(\dfrac{60}{2}\) = 1 . 3 . 5.... 59
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)
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BT5 : Chứng minh
2) \(\dfrac{3}{5}< \dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+....+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{4}{5}\)
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)
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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
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So sánh \(P=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}\) và Q=1.3.5.7........59
Ta có:
31/2.32/2.33/2....60/2=31.32......60/2^30
=(31.32.33....60)(1.2.3....30)/2^30(1.2.3...30)
=(1.3.5...59)(2.4.6...60)/(2.4.6...60)=1.3.5...59
=>P=Q
nhớ ****
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cái dòng 3, 4 mk ko hiểu sao 2^30.(1.2.3....30) lại bằng 2.4.6...60
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Xem thêm câu trả lời
so sánh:
\(P=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}\)và \(Q=1.3.5.7...59\)
\(\dfrac{31}{2}\cdot\dfrac{32}{2}...\dfrac{60}{2}=1.3.5...59\)
so sánh \(P=\frac{31}{2}.\frac{32}{2}.\frac{33}{2}...\frac{60}{2}\) và \(Q=1.3.5.7...59\)
So sánh Q, P biết :
P = \(\dfrac{20}{30}\) + \(\dfrac{20}{70}\) + \(\dfrac{20}{126}\) + ... + \(\dfrac{20}{798}\)
Q = (\(\dfrac{31}{2}\) . \(\dfrac{32}{2}\) . \(\dfrac{33}{2}\) ... \(\dfrac{60}{2}\)) : ( 1.3.5...55)
So sánh S=\(\dfrac{1}{2}\)+\(\dfrac{1}{11}\)+\(\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}\)với \(\dfrac{9}{10}\)
Ta có: \(S< \dfrac{1}{2}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{11}+\dfrac{1}{31}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{32}\) \(=\dfrac{1}{2}+\dfrac{3}{11}+\dfrac{2}{31}+\dfrac{2}{32}\)
\(=\dfrac{4909}{5456}< \dfrac{9}{10}\)
\(\Rightarrow S< \dfrac{9}{10}\)
Vậy \(S< \dfrac{9}{10}\)
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So sánh: A=\(\dfrac{1}{2}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}\)với 0,9
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