Chứng minh :
A =\(\dfrac{1}{2}+\dfrac{1}{33}+\dfrac{1}{34}+\dfrac{1}{35}+\dfrac{1}{51}+\dfrac{1}{53}+\dfrac{1}{55}+\dfrac{1}{57}+\dfrac{1}{59}\)<\(\dfrac{7}{10}\)
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NH
5 tháng 3 2023
\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)
\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)
5 tháng 3 2023
Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)
\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)
\(\Leftrightarrow\left(x+60\right)\text{=}0\)
\(x\text{=}-60\)
\(Vậy...\)
13 tháng 3 2017
Đặt \(A=\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
Ta có:
\(A=\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\)
\(\Rightarrow A=\frac{1}{3}+\left(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}\right)+\left(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}\right)\)
Nhận xét:
\(\frac{1}{31}+\frac{1}{35}+\frac{1}{37}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}\)
\(\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{45}+\frac{1}{45}+\frac{1}{45}\)
\(\Rightarrow A< \frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)\)
Mà \(\frac{1}{3}+\left(\frac{1}{30}+\frac{1}{30}+\frac{1}{30}\right)+\left(\frac{1}{45}+\frac{1}{45}+\frac{1}{45}\right)=\frac{1}{3}+\frac{1}{10}+\frac{1}{15}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}< \frac{1}{2}\) (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
20 tháng 6 2017
a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$
$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$
$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$
$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$
b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$
$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$
$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$
DT
Đỗ Thanh Hải
CTVVIP
10 tháng 3 2021
Có thể làm như sau
Ta thấy \(\dfrac{1}{51}< \dfrac{1}{50}\)
\(\dfrac{1}{52}< \dfrac{1}{50}\)
.......
\(\dfrac{1}{100}< \dfrac{1}{50}\)
=> A = \(\dfrac{1}{50}+\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}.50=1\)
Lại có
\(\dfrac{1}{51}>\dfrac{1}{100}\)
\(\dfrac{1}{52}>\dfrac{1}{100}\)
.......
\(\dfrac{1}{99}>\dfrac{1}{100}\)
=> A = \(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\)
=> \(\dfrac{1}{2}< A< 1\)
Vậy A không phải số tự nhiên
18 tháng 3 2017
Ta thấy:
\(\dfrac{1}{51}< \dfrac{1}{50}\)
\(\dfrac{1}{52}< \dfrac{1}{50}\)
...
\(\dfrac{1}{100}< \dfrac{1}{50}\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< \dfrac{1}{50}.50=1\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< 1\left(1\right)\)
Lại có:
\(\dfrac{1}{51}>\dfrac{1}{100}\)
\(\dfrac{1}{52}>\dfrac{1}{100}\)
...
\(\dfrac{1}{100}=\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}>\dfrac{1}{100}.50=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}>\dfrac{1}{2}\left(2\right)\)
Từ (1),(2)\(\Rightarrow\)\(\dfrac{1}{2}< \dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}< 1\)
12 tháng 7 2017
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\left(n\in N\right)\)
\(B=\dfrac{10^{20}+1}{10^{21}+1}< 1\)
\(B< \dfrac{10^{20}+1+9}{10^{21}+1+9}\Rightarrow B< \dfrac{10^{20}+10}{10^{21}+10}\Rightarrow B< \dfrac{10\left(10^{19}+1\right)}{10\left(10^{20}+1\right)}\Rightarrow B< \dfrac{10^{19}+1}{10^{20}+1}=A\)\(\Rightarrow B< A\)
DN
1
21 tháng 8 2021
Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}-\dfrac{x-60}{56}-\dfrac{x-60}{55}-\dfrac{x-60}{54}=0\)
\(\Leftrightarrow x-60=0\)
hay x=60
2 tháng 11 2021
Bài 1:
1) \(9A=3^3+3^5+...+3^{113}\)
\(\Rightarrow8A=9A-A=3^3+3^5+...+3^{113}-3-3^3-...-3^{111}=3^{113}-3\)
\(\Rightarrow A=\dfrac{3^{113}-3}{8}\)
2) \(9B=3^4+3^6+...+3^{202}\)
\(\Rightarrow8B=9B-B=3^4+3^6+...+3^{202}-3^2-3^4-...-3^{200}=3^{202}-3^2=3^{202}-9\)
\(\Rightarrow B=\dfrac{3^{202}-9}{8}\)
3) \(25C=5^3+5^5+...+5^{101}\)
\(\Rightarrow24C=25C-C=5^3+5^5+...+5^{101}-5-5^3-...-5^{99}=5^{101}-5\)
\(\Rightarrow C=\dfrac{5^{101}-5}{24}\)
4) \(25D=5^4+5^6+...+5^{102}\)
\(\Rightarrow24D=25D-D=5^4+5^6+...+5^{102}-5^2-5^4-...-5^{100}=5^{102}-25\)
\(\Rightarrow D=\dfrac{5^{102}-25}{24}\)
2 tháng 11 2021
Bài 2:
a) Gọi d là UCLN(2n+1,n+1)
\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\n+1⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}2n+1⋮d\\2n+2⋮d\end{matrix}\right.\)
\(\Rightarrow\left(2n+2\right)-\left(2n+1\right)⋮d\Rightarrow1⋮d\)
Vậy 2n+1 và n+1 là 2 số nguyên tố cùng nhau
\(\Rightarrow\dfrac{2n+1}{n+1}\) là phân số tối giản
b) Gọi d là UCLN(2n+3,3n+4)
\(\Rightarrow\left\{{}\begin{matrix}2n+3⋮d\\3n+4⋮d\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}6n+9⋮d\\6n+8⋮d\end{matrix}\right.\)
\(\Rightarrow\left(6n+9\right)-\left(6n+8\right)⋮d\Rightarrow1⋮d\)
\(\Rightarrow\dfrac{2n+3}{3n+4}\) là phân số tối giản
Lời giải:
\(A=\frac{1}{2}+\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}\)
Ta có:
\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}< \frac{1}{30}+\frac{1}{30}+\frac{1}{30}=\frac{3}{30}=\frac{1}{10}\)
\(\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}< \frac{1}{50}+\frac{1}{50}+\frac{1}{50}+\frac{1}{50}+\frac{1}{50}=\frac{5}{50}=\frac{1}{10}\)
Cộng theo vế:
\(\frac{1}{33}+\frac{1}{34}+\frac{1}{35}+\frac{1}{51}+\frac{1}{53}+\frac{1}{55}+\frac{1}{57}+\frac{1}{59}< \frac{2}{10}=\frac{1}{5}\)
Suy ra \(A< \frac{1}{2}+\frac{1}{5}=\frac{7}{10}\)
Ta có đpcm.