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Ta đặt cm là A
Vì 1/2 < 2/3 ; 3/4 < 4/5 ; 5/6 < 6/7 ; ...;99/100<100/101
=> A = 1/2 x 3/4 x 5/6 x...x 99/100 < B= 2/3 X 4/5 X 6/7 X....X100/101
=> A x A < A x B = 1 x 3 x 5 x 99 / 2 x 4 x 6 x ......x 100 x 2 x 4 x 6 x ...x 100/3 x 5 x 7 x ...x 101
Ta rút gọn 2 x 4 x 6 x ..x 100 và 3 x 5 x ...x 99 ta còn 1/101
=>A^2 < 1/101 => A^2 < 1/101 < 1/100 = > A ^ 2 <1/100 => A^2 ,(1/10 ^2
=> A < 1/10
Chứng minh A > 1/15
1/2 = 1/2
3/4 >2/3
5/6 > 4/5
......
99/100 > 98/99
A^2 > 1/2 x ( 1/2 x 2/3 x 3/4 x ...x 98/99 x 99/100
A^2 > 1/2 x 1/100
A^2 > 1/200 > 1/225
A^2 > (1/15) ^2
Vậy A > 1/15
a) \(\frac{2}{3}x>-6\)
=> \(x>\left(-6\right):\frac{2}{3}\)
=> \(x>-9\)
b) \(-\frac{5}{6}x< 20\)
=> \(x< 20:-\frac{5}{6}\)
=> \(x>-24\)
c) \(3-\frac{1}{4}x>2\)
=> \(\frac{1}{4}x< 3-2\)
=> \(\frac{1}{4}x< 1\)
=> \(x< 4\)
d) \(5-\frac{1}{3}x>3\)
=> \(\frac{1}{3}x< 5-3\)
=> \(\frac{1}{3}x< 2\)
=> \(x< 6\)
a) \(\frac{x+5}{4}\)-\(\frac{2x-5}{3}\)=\(\frac{6x-1}{3}\)+\(\frac{2x-3}{12}\)
⇔\(\frac{3\left(x+5\right)}{12}\)-\(\frac{4\left(2x-5\right)}{12}\)=\(\frac{4\left(6x-1\right)}{12}\)+\(\frac{2x-3}{12}\)
⇒ 3x+15-8x+20=24x-4+2x-3
⇔3x+15-8x+20-24x+4-2x+3=0
⇔-31x+42=0
⇔x=\(\frac{42}{31}\)
Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{42}{31}\)}
Ta có:
\(\left(a+b-c\right)^2\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge2ac+2bc-2ab\)
Mà \(a^2+b^2+c^2=\frac{5}{3}< 2\)
\(\Rightarrow2ac+2bc-2ab< 2\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}\)
A=1/2^2 + 1/3^2 + 1/4^2 + ... + 1/2017^2
A < 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/2016.2017
A < 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2016 - 1/2017
A < 1 - 1/2017 < 1 (1)
B = 2!/3! + 2!/4! + 2!/5! + ... + 2!/2017!
B = 2!.(1/3! + 1/4! + 1/5! + ... + 1/2017!)
B < 2.(1/2.3 + 1/3.4 + 1/4.5 + ... + 1/2016.2017)
B < 2.(1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/2016 - 1/2017)
B < 2.(1/2 - 1/2017) < 2.1/2 = 1 (2)
Từ (1) và (2) => A + B < 2 (đpcm)
\(a.\frac{x+5}{2021}+\frac{x+6}{2020}+\frac{x+7}{2019}=-3\\ \Leftrightarrow\frac{x+5}{2021}+1+\frac{x+6}{2020}+1+\frac{x+7}{2019}+1=0\\ \Leftrightarrow\frac{x+2026}{2021}+\frac{x+2026}{2020}+\frac{x+2026}{2019}=0\\ \Leftrightarrow\left(x+2026\right)\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\right)=0\\\Leftrightarrow x+2026=0\left(Vi\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}\ne0\right)\\ \Leftrightarrow x=-2026\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-2026\right\}\)
\(b.\frac{2-x}{100}-1=\frac{1-x}{101}-\frac{x}{102}\\ \Leftrightarrow\frac{2-x}{100}+1=\frac{1-x}{101}+1+1-\frac{x}{102}\\\Leftrightarrow \frac{102-x}{100}-\frac{102-x}{101}-\frac{102-x}{102}=0\\ \Leftrightarrow\left(102-x\right)\left(\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\right)=0\\ \Leftrightarrow102-x=0\left(Vi\frac{1}{100}-\frac{1}{101}-\frac{1}{102}\ne0\right)\\ \Leftrightarrow x=102\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{102\right\}\)
c/ PT tương đương
\(\frac{x+1}{93}-1+\frac{x-2}{45}-2+\frac{x+4}{32}-3=0\)
\(\Leftrightarrow\frac{x-92}{93}+\frac{x-92}{45}+\frac{x-92}{32}=0\)
\(\Leftrightarrow\left(x-92\right)\left(\frac{1}{93}+\frac{1}{45}+\frac{1}{32}\right)=0\Rightarrow x=92\)
_Appreciate:
\(3^2=2.4+1\)
\(5^2=4.6+1\)
...
\(\left(2n+1\right)^2=2n\left(2n+2\right)+1\)
_Solution:
\(A=\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+...+\frac{1}{\left(2n+1\right)^2}< \frac{1}{3^2-1}+\frac{1}{5^2-1}+...+\frac{1}{\left(2n+1\right)^2-1}\)
\(A< \frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{2n.\left(2n+2\right)}\)\(A< \frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2n}-\frac{1}{2n+2}\right)\)
\(A< \frac{1}{2}.\left(\frac{1}{2}-\frac{1}{2n+2}\right)=\frac{1}{4}-\frac{1}{2.\left(2n+2\right)}< \frac{1}{4}\) (proof)
Giải
\(S_{\infty} = \sum_{k = 1}^{\infty} \frac{2 k - 1}{3^{2 k}} .\)
Khi đó \(A\) là tổng hữu hạn 50 số hạng đầu, nên rõ ràng \(A < S_{\infty}\).
\(\sum_{k = 1}^{\infty} k \textrm{ } x^{k} = \frac{x}{\left(\right. 1 - x \left.\right)^{2}} , \sum_{k = 1}^{\infty} x^{k} = \frac{x}{1 - x} .\)
\(\sum_{k = 1}^{\infty} \left(\right. 2 k - 1 \left.\right) \textrm{ } x^{k} = 2 \sum_{k = 1}^{\infty} k \textrm{ } x^{k} \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \sum_{k = 1}^{\infty} x^{k} = 2 \cdot \frac{x}{\left(\right. 1 - x \left.\right)^{2}} \textrm{ }\textrm{ } - \textrm{ }\textrm{ } \frac{x}{1 - x} = \frac{2 x}{\left(\right. 1 - x \left.\right)^{2}} - \frac{x}{1 - x} .\)
\(\frac{2 x}{\left(\right. 1 - x \left.\right)^{2}} = \frac{2 x}{\left(\right. 1 - x \left.\right)^{2}} , \frac{x}{1 - x} = \frac{x \left(\right. 1 - x \left.\right)}{\left(\right. 1 - x \left.\right)^{2}} = \frac{x - x^{2}}{\left(\right. 1 - x \left.\right)^{2}} .\)
Suy ra
\(\sum_{k = 1}^{\infty} \left(\right. 2 k - 1 \left.\right) \textrm{ } x^{k} = \frac{2 x - \left(\right. x - x^{2} \left.\right)}{\left(\right. 1 - x \left.\right)^{2}} = \frac{2 x - x + x^{2}}{\left(\right. 1 - x \left.\right)^{2}} = \frac{x + x^{2}}{\left(\right. 1 - x \left.\right)^{2}} .\)
\(S_{\infty} = \frac{\frac{1}{9} + \frac{1}{81}}{\left(\right. 1 - \frac{1}{9} \left.\right)^{2}} = \frac{\frac{1}{9} + \frac{1}{81}}{\left(\right. \frac{8}{9} \left.\right)^{2}} = \frac{\frac{9 + 1}{81}}{\frac{64}{81}} = \frac{\frac{10}{81}}{\frac{64}{81}} = \frac{10}{64} = \frac{5}{32} .\)
\(A < S_{\infty} = \frac{5}{32} .\)
Vậy đã chứng minh \(A < \frac{5}{32} .\)