1+23+9999999999999999999989999999999999
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\(B=\frac{23^{41}+1}{23^{42}+1}\)
Vì B < 1
\(\Rightarrow B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23(23^{40}+1)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)
P/s: Hoq chắc
ta có
\(B=\frac{23^{41}+1}{23^{42}+1}< \frac{23^{41}+1+22}{23^{42}+1+22}=\frac{23^{41}+23}{23^{42}+23}=\frac{23\left(23^{40}+1\right)}{23\left(23^{41}+1\right)}=\frac{23^{40}+1}{23^{41}+1}=A\)
\(\Rightarrow B< A\)
\(1(đvC)=\dfrac{1}{12}.1,9926.10^{-23}=1,6605.10^{-24}(g)\\ \Rightarrow m_{Al}=27(đvC)=27.1,6605.10^{-24}\approx 4,48.10^{-23}(g)\)
Chọn C
Bn hâm mộ lewandoski à? Ghi 41 bàn ở mùa giải 2020 - 2021 đó
a) 23x51+75x23-23x25=23x(51+75-26)=23x100=2300
b) 1+2+3+...+20=\(\dfrac{\left(20+1\right)\text{x}20}{2}\)=210
a) \(23\times51+75\times23-23\times26\)
\(=23\times\left(51+75-26\right)\)
\(=23\times\left(126-26\right)\)
\(=23\times100\)
\(=2300\)
b) \(1+2+...+20\)
\(=\left(20+1\right)+\left(19+2\right)+\left(18+3\right)+\left(17+4\right)+...+\left(11+10\right)\)
\(=21+21+21+...+21\) (10 số 21)
\(=2100\)
10000000000000000000000000000000023
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