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1) \(23^{401}+38^{202}-2^{433}=23^{4.100}.23+38^{4.50}.38^2-2^{4.108}.2^1=\left(..1\right).23+\left(..6\right).1444-\left(..6\right).2=\left(..3\right)+\left(..4\right)-\left(..2\right)=\left(..5\right)\)
Set \(S=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2023}}\)
Then \(3S=1+\dfrac{1}{3}+...+\dfrac{1}{3^{2022}}\)
Hence \(2S=3S-S=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{2022}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2023}}\right)\)
\(=1-\dfrac{1}{3^{2023}}\)
\(\Leftrightarrow S=\dfrac{1}{2}-\dfrac{1}{2.3^{2023}}< \dfrac{1}{2}\) (Q. E. D)
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\)
Ta có: \(3A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\)
\(3A-A=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{2022}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2023}}\right)\)
\(2A=1-\dfrac{1}{3^{2023}}\)
\(A=\dfrac{1-\dfrac{1}{3^{2023}}}{2}\)
Vì \(\dfrac{1-\dfrac{1}{3^{2023}}}{2}< \dfrac{1}{2}\) nên \(A< \dfrac{1}{2}\)
Vậy...
A=1/3+1/3^2+...+1/3^2005
=> 3A= 1+1/3+...+1/3^2004
=> 3A-A=(1+1/3+...+1/3^2004)-(1/3+1/3^2+...+1/3^2005)
=> 2A =1-1/3^2005 <1
=> A<1/2
Ta có \(a^2>a^2-1\forall a\)
\(\Rightarrow a^2>\left(a-1\right)\left(a+1\right)\)
\(\Rightarrow\dfrac{1}{a^2}< \dfrac{1}{\left(a-1\right)\left(a+1\right)}=\dfrac{1}{2}\cdot\left(\dfrac{1}{a-1}\right)\left(\dfrac{1}{a+1}\right)\)
Áp dụng, ta có
\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1+\dfrac{1}{2^2}+\dfrac{1}{2\cdot4}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)
= \(1+\dfrac{1}{2^2}+\dfrac{1}{2}\cdot\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)
= 1+ \(\dfrac{1}{4}\)+\(\dfrac{1}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
=1+ \(\dfrac{2}{3}-\dfrac{1}{2}\cdot\left(\dfrac{1}{n}+\dfrac{1}{n+1}\right)\) < \(1+\dfrac{2}{3}=\dfrac{5}{3}\left(ĐPCM\right)\)
(Mik mượn chỗ bình luận ké nha!!)
Người Ấy Là Ai-eqt đẹp đó :)
cmr: 1+2=3
3-2x1=1
vậy crm= 1
giả sử:em có 1 cái kẹo, anh có 2 cái kẹo
tính tay 1 nhón tay và hai ngón tay , đếm 1;2;3