\(ChoB=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\). Hãy chứng tỏ rằng \(B>1\)
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NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2022
\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{20}-\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{20}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{20}-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{10}\right)\)
\(=\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{20}\) (đpcm)
PN
3
25 tháng 4 2017
Ta có :
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...........+\dfrac{1}{19}\)
\(B=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+.......+\dfrac{1}{19}\right)\)
Ta thấy :
\(\dfrac{1}{5}>\dfrac{1}{20}\)
\(\dfrac{1}{6}>\dfrac{1}{20}\)
..................
\(\dfrac{1}{19}>\dfrac{1}{20}\)
\(\Rightarrow B>\dfrac{1}{4}+\left(\dfrac{1}{20}+\dfrac{1}{20}+.........+\dfrac{1}{20}\right)\)(\(15\) p/s \(\dfrac{1}{20}\))
\(B>\dfrac{1}{4}+\dfrac{1}{20}.15\)
\(B>\dfrac{1}{4}+\dfrac{3}{4}=1\Rightarrow B>1\rightarrowđpcm\)
~ Học tốt ~
DT
Đỗ Thanh Hải
CTVVIP
2 tháng 5 2021
Ta có
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...............
\(\dfrac{1}{8^2}< \dfrac{1}{7.8}\)
=> B < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{7.8}\)
B < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
B < \(1-\dfrac{1}{8}< 1\) (Do \(\dfrac{1}{8}>0\))
Vậy.....
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
a)
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{30^2}\\ < \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{29.30}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{29}-\dfrac{1}{30}\\ =1-\dfrac{1}{30}=\dfrac{29}{30}< 1\left(dpcm\right)\)
b)
\(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}=\dfrac{1}{10}+\left(\dfrac{1}{11}+\dfrac{1}{12}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\\ >\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{1}{10}+\dfrac{90}{100}\\ =\dfrac{110}{100}>1\left(đpcm\right).\)
NA
Ngoc Anh Thai
Giáo viên
11 tháng 4 2021
c)
\(\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+...+\dfrac{1}{17}\\ =\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{9}\right)+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{17}\right)\\ < \dfrac{1}{5}.5+\dfrac{1}{8}.8=1+1=2\left(đpcm\right)\)
d) tương tự câu 1
12 tháng 4 2017
\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)
\(B< 1-\dfrac{1}{8}=\dfrac{7}{8}< 1\)
mink nhanh nhất đó bạn,
4 tháng 5 2018
ta có :
\(\dfrac{1}{2^2}< \dfrac{1}{1\times2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\times3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\times4}\)
. . . . . . .
\(\dfrac{1}{8^2}< \dfrac{1}{7\times8}\)
_________________________________
\(\Rightarrow\)\(B< \)\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{7.8}\right)\)
\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{7}-\dfrac{1}{8}\)
\(\Rightarrow B< 1-\dfrac{1}{8}\)
\(\Rightarrow B< 1\)
\(\Rightarrowđpcm\)
NA
Nguyễn An Ninh
VIP
25 tháng 4 2023
b\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 3/4
NA
Nguyễn An Ninh
VIP
25 tháng 4 2023
Tương tự như vậy với câu a\()\)
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2.3 + 1/3.4 +... + 1/99.100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/3 + 1/3 -1/4 +... + 1/99 + 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 1/4 + 1/2 - 1/100
1/2^2 + 1/3^2 +... + 1/100^2 < 3/4 - 1/100 < 1/2
14 tháng 2 2023
Ta có :
\(\dfrac{1}{2}>\dfrac{1}{5}\)
\(\dfrac{1}{3}>\dfrac{1}{5}\)
\(\dfrac{1}{4}>\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\)
YN
14 tháng 2 2023
\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{7}{6}+\dfrac{1}{4}+\dfrac{1}{5}\)
\(=\dfrac{17}{12}+\dfrac{1}{5}\)
\(=\dfrac{97}{60}\)
\(\dfrac{4}{5}=\dfrac{4.12}{5.12}=\dfrac{48}{60}\)
Mà \(\dfrac{97}{60}>\dfrac{48}{60}\)
\(\Rightarrow\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}>\dfrac{4}{5}\left(đpcm\right)\).
\(B=\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\)
\(=\dfrac{1}{4}+\left(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\right)\)
Các phân số \(\dfrac{1}{5}\), \(\dfrac{1}{6}\), \(\dfrac{1}{7}\), ..., \(\dfrac{1}{19}\) đều lớn hơn \(\dfrac{1}{20}\), tất cả có 15 phân số nên:
\(B>\dfrac{1}{4}+\left(\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\right)=\dfrac{1}{4}+\dfrac{3}{4}=1\)
Vậy B > 1
e! Chung minh di tai sao lai lam the : phai co ly do chu( ko phai cu thich la ko lam ngay duoc dau