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\(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^7}\)
\(=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^7}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^7}\right)\)
\(=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+...+\dfrac{1}{2^6}-\dfrac{1}{2^1}-\dfrac{1}{2^2}-...-\dfrac{1}{2^7}\)
\(=1-\dfrac{1}{2^7}\)
\(=\dfrac{127}{128}\)
A = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
A x 2 = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
A x 2 - A = 1 + \(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+\(\dfrac{1}{64}\) - (\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{128}\))
A x (2 - 1) = 1+\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)+\(\dfrac{1}{8}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{32}\)+\(\dfrac{1}{64}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)-\(\dfrac{1}{8}\)-\(\dfrac{1}{16}\)-\(\dfrac{1}{32}\)-\(\dfrac{1}{64}\)-\(\dfrac{1}{128}\)
A = (1 - \(\dfrac{1}{128}\)) +(\(\dfrac{1}{2}\)-\(\dfrac{1}{2}\)) + (\(\dfrac{1}{4}\) - \(\dfrac{1}{4}\)) +...+(\(\dfrac{1}{64}\) - \(\dfrac{1}{64}\))
A = 1 - \(\dfrac{1}{128}\)
A = \(\dfrac{127}{128}\)
Ta có: \(\left\{{}\begin{matrix}\left(x-2\right)^4\ge0\forall x\\\left(2y-1\right)^{2022}\ge0\forall y\end{matrix}\right.\)
\(\Rightarrow\left(x-2\right)^4+\left(2y-1\right)^{2022}\ge0\forall x,y\)
Mà: \(\left(x-2\right)^4+\left(2y-1\right)^{2022}\le0\)
Do đó: \(\left(x-2\right)^4+\left(2y-1\right)^{2022}=0\)
Khi đó: \(\left\{{}\begin{matrix}x-2=0\\2y-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)
Thay \(x=2;y=\dfrac{1}{2}\) vào M, ta được:
\(M=21\cdot2\cdot\left(\dfrac{1}{2}\right)^2+4\cdot2\cdot\left(\dfrac{1}{2}\right)^2\)
\(=25\cdot2\cdot\left(\dfrac{1}{2}\right)^2=\dfrac{25}{2}\)
\(\text{#}Toru\)
(\(x\) - 2)4 + (2y - 1)2022 ≤ 0
Vì: ( \(x-2\))4 ≥ 0 \(\forall\) \(x\); (2y - 1)2022 ≥ 9 \(\forall\) y
Vậy (\(x-2\))4 + (2y - 1)2022 = 0
⇒ \(\left\{{}\begin{matrix}x-2=0\\2y-1=0\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=2\\2y=1\end{matrix}\right.\)
⇒ \(\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\) (1)
Thay hệ (1) vào biểu thức M = 21\(xy^2\) + 4\(xy^2\)
M = 21.2.\(\dfrac{1}{2^2}\) + 4.2.\(\dfrac{1}{2^2}\)
M = 2.\(\dfrac{1}{2^2}\).(21 + 4)
M = \(\dfrac{1}{2}\).25
M = \(\dfrac{25}{2}\)
1.3\(x-1\) + 5.3\(x-1\) = 162
3\(^{x-1}\).(1 + 5) = 162
3\(x-1\).6 = 162
3\(x-1\) = 162 : 6
3\(^{x-1}\) = 27
3\(^{x-1}\) = 33
\(x-1\) = 3
\(x\) = 3 + 1
\(x\) = 4
Vậy \(x=4\)
Lời giải:
$(x+1)+(x+2)+(x+3)+....+(x+99)+(x+100)=5050$
$(x+x+....+x)+(1+2+3+...+100)=5050$
Số lần xuất hiện của $x$: $(100-1):1+1=100$ (lần)
Suy ra:
$x\times 100+(1+2+...+100)=5050$
$x\times 100+100\times 101:2=5050$
$x\times 100+5050=5050$
$x\times 100=0$
$x=0:100$
$x=0$
Ta thấy:\(\dfrac{16}{18}=\dfrac{32}{36\dfrac{ }{ }}=\dfrac{48}{54}=\dfrac{64}{72}=\dfrac{70}{90}=\dfrac{ }{ }\)
Vậy là có 4 P/s =16/18 có cả tử và mẫu có 2 chữ số
