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Hơi nhầm xíu 113 . 7^2+8^2=113 cứ tưởng 112. Hơi ngáo tí =[[
Lời giải
Biến đổi tương đương ta được: \(L=\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{25}+\dfrac{1}{41}+\dfrac{1}{61}+\dfrac{1}{85}+\dfrac{1}{113}=\dfrac{1}{1^2+2^2}+\dfrac{1}{2^2+3^2}+\dfrac{1}{3^2+4^2}+\dfrac{1}{4^2+5^2}+\dfrac{1}{5^2+6^2}+\dfrac{1}{6^2+7^2}+\dfrac{1}{7^2+8^2}\)
\(L=\dfrac{1}{1^2+\left(1+1\right)^2}+\dfrac{1}{2^2+\left(2+1\right)^2}+...+\dfrac{1}{7^2+\left(7+1\right)^2}\)
Chứng minh 1 bđt cơ bản sau: \(n^2+\left(n+1\right)^2>2n\left(n+1\right)\) thật vậy:
\(n^2+\left(n+1\right)^2=n^2+n^2+2n+1=2n^2+2n+1=2n\left(n+1\right)+1>2n\left(n+1\right)\)
\(\Rightarrow\dfrac{1}{n^2+\left(n+1\right)^2}< \dfrac{1}{2n\left(n+1\right)}\)
trở lại bài toán ta có: \(L< \dfrac{1}{2.1.2}+\dfrac{1}{2.2.3}+...+\dfrac{1}{2.7.8}\)
\(L< \dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..+\dfrac{1}{7.8}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+..+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{8}\right)=\dfrac{1}{2}-\dfrac{1}{16}< \dfrac{1}{2}\left(đpcm\right)\)
Đề sai đúng hk? CHỗ kia 112 chứ lấy đâu ra 113
p/s : 7^2+8^2=112. =))
bài 2
| a;đặt biểu thức là S | |
| S < 1/1.2 + 1/2.3 + .......1/(n-1)n | |
| = 1- 1/2 +1 /2 -1/3+........ + 1/n-1 - 1/n | |
|
= 1 -1/n <1 |
|
| vậy S < 1 | |
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
Các câu còn lại tương tự
a) \(A=2A-A\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2021}}-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2022}}\right)\)
\(=1-\dfrac{1}{2^{2022}}\)
b) \(B=\dfrac{20+15+12+17}{60}=\dfrac{4}{5}=1-\dfrac{1}{5}\)
\(A>B\left(Vì\left(\dfrac{1}{2^{2022}}< \dfrac{1}{5}\right)\right)\)
2/
a/ \(\dfrac{7}{10}=\dfrac{7.15}{10.15}=\dfrac{105}{150}\)
\(\dfrac{11}{15}=\dfrac{11.10}{15.10}=\dfrac{110}{150}\)
-Vì \(\dfrac{105}{150}< \dfrac{110}{150}\)(105<110)nên \(\dfrac{7}{10}< \dfrac{11}{15}\)
b/ \(\dfrac{-1}{8}=\dfrac{-1.3}{8.3}=\dfrac{-3}{24}\)
-Vì \(\dfrac{-3}{24}>\dfrac{-5}{24}\left(-3>-5\right)\)nên\(\dfrac{-1}{8}>\dfrac{-5}{24}\)
c/\(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)
\(\dfrac{10}{40}=\dfrac{10:10}{40:10}=\dfrac{1}{4}\)
-Vì \(\dfrac{1}{4}=\dfrac{1}{4}\)nên\(\dfrac{25}{100}=\dfrac{10}{40}\)
a/ \(\dfrac{7}{10}< \dfrac{11}{15}\)
c/ \(\dfrac{25}{100}=\dfrac{10}{40}\)
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B
Lời giải:
Ta có:
\(\left\{\begin{matrix} \frac{1}{13}< \frac{1}{12}\\ \frac{1}{14}< \frac{1}{12}\\ \frac{1}{15}< \frac{1}{12}\end{matrix}\right.\Rightarrow \frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{3}{12}=\frac{1}{4}(1)\)
\(\left\{\begin{matrix} \frac{1}{61}< \frac{1}{60}\\ \frac{1}{62}< \frac{1}{60}\\ \frac{1}{63}< \frac{1}{60}\end{matrix}\right.\Rightarrow \frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{3}{60}=\frac{1}{20}(2)\)
Từ \((1);(2)\Rightarrow \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}\)
Hay \( \frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)
Ta có đpcm.
Đặt A là biểu thức đó
Ta có:
\(\dfrac{1}{13}< \dfrac{1}{12};\dfrac{1}{14}< \dfrac{1}{12};\dfrac{1}{15}< \dfrac{1}{12}\)
\(\Rightarrow\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}\)
Ta cũng có
\(\dfrac{1}{61}< \dfrac{1}{60};\dfrac{1}{62}< \dfrac{1}{60};\dfrac{1}{63}< \dfrac{1}{60}\)
\(\Rightarrow\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)
\(\Rightarrow A< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
\(\Rightarrow\)dpcm
\(A=\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)
\(A=\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=3.\dfrac{1}{3}.\left(1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)\)
\(\Rightarrow3A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)
\(\Rightarrow3A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\left(\dfrac{1}{3^1}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow2A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...\dfrac{1}{3^{2022}}-\dfrac{1}{3^1}-\dfrac{1}{3^2}-\dfrac{1}{3^3}-...\dfrac{1}{3^{2022}}-\dfrac{1}{3^{2023}}\)
\(\Rightarrow2A=1-\dfrac{1}{3^{2023}}\)
\(\Rightarrow A=\dfrac{1}{2}\left(1-\dfrac{1}{3^{2023}}\right)\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{2023}}< \dfrac{1}{2}\)
\(B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4+3+1}{12}=\dfrac{8}{12}=\dfrac{2}{3}\)
mà \(\dfrac{2}{3}>\dfrac{1}{2}\) \(\left(\dfrac{2}{3}=\dfrac{4}{6}>\dfrac{1}{2}=\dfrac{3}{6}\right)\)
\(\Rightarrow A< B\)
A = \(\dfrac{1}{3}\)+ \(\dfrac{1}{3^2}\)+ \(\dfrac{1}{3^3}\)+............+\(\dfrac{1}{3^{2023}}\)
3A = 1+ \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^3}\)+...+\(\dfrac{1}{3^{2022}}\)
3A - A = 1 - \(\dfrac{1}{3^{2023}}\)
2A = 1 - \(\dfrac{1}{3^{2023}}\) < 1
B = \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{12}\)
B = \(\dfrac{4}{12}\) + \(\dfrac{3}{12}\) + \(\dfrac{1}{12}\)
B = \(\dfrac{8}{12}\)
B = \(\dfrac{2}{3}\) ⇒ 2B = \(\dfrac{4}{3}\) > 1
2A < 2B ⇒ A < B
