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mik mới so sánh được với 98 thôi. Mik ra kq là A > 99 - 99/100 -> A > 98 chứ chưa so sánh đc với 99.
P = (1-1/2).(1-1/3).(1-1/4)...(1-1/99) = 1/2 . 2/3 . 3/4 ... 98/99 = 1/99
Ta có :
\(P=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).................\left(1-\dfrac{1}{99}\right)\)
\(P=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)...............\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)
\(P=\dfrac{1}{2}.\dfrac{2}{3}..................\dfrac{98}{99}\)
\(P=\dfrac{1}{99}\)
~ Học tốt ~
\(T=\left(\dfrac{1}{2}+1\right).\left(\dfrac{1}{3}+1\right).\left(\dfrac{1}{4}+1\right).......\left(\dfrac{1}{98}+1\right).\left(\dfrac{1}{99}+1\right) \) \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}....\dfrac{99}{98}\cdot\dfrac{100}{99}\)
\(=\dfrac{100}{2}=50\)
\(T=\left|\dfrac{1}{2}+1\right|\left|\dfrac{1}{3}+1\right|\left|\dfrac{1}{4}+1\right|.....\left|\dfrac{1}{98}+1\right|\left|\dfrac{1}{99}+1\right|\)
\(T=\left|\dfrac{3}{2}\right|.\left|\dfrac{4}{3}\right|.\left|\dfrac{5}{4}\right|......\left|\dfrac{99}{98}\right|.\left|\dfrac{100}{99}\right|\)
\(T=\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.....\dfrac{99}{98}.\dfrac{100}{99}\)
\(T=\dfrac{3.4.5.....99.100}{2.3.4.....98.99}=\dfrac{100}{2}=50\)
\(M=2^{100}-2^{99}+2^{98}-2^{97}+...+2-1\)
\(\Rightarrow2M=2\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2-1\right)\)
\(2M=2^{101}-2^{100}+2^{99}-2^{98}+...+2^2-2\)
\(2M+M=3M=2^{101}-2^{100}+2^{99}-2^{98}+...+2^2-2+2^{100}-2^{99}+2^{98}-2^{97}+...+2-1\)
\(3M=2^{101}-1\Leftrightarrow M=\dfrac{2^{101}-1}{3}\) vậy \(M=\dfrac{2^{101}-1}{3}\)
\(M=2^{100}-2^{99}+2^{98}-2^{97}+...+2-1\)
\(2M=2\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2-1\right)\)
\(2M=2^{101}-2^{100}+2^{99}-2^{98}+...+2^2-2\)
\(2M+M=\left(2^{101}-2^{100}+2^{99}-2^{98}+...+2^2-2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+...+2-1\right)\)
\(3M=2^{101}-1\)
\(M=\dfrac{2^{101}-1}{3}\)
Ta có: \(\left|x-y\right|+\left|x-1\right|\ge0\)
\(\Rightarrow A=\left|x-y\right|+\left|x-1\right|+2017\ge2017\)
Dấu " = " khi \(\left\{{}\begin{matrix}\left|x-y\right|=0\\\left|x-1\right|=0\end{matrix}\right.\Rightarrow x=y=1\)
Vậy \(MIN_A=2017\) khi x = y = 1
a, Ta có: \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{81}\right)^7=\left(\dfrac{1}{3^4}\right)^7=\left(\dfrac{1}{3}\right)^{28}=\dfrac{1}{3^{28}}\)
\(\left(\dfrac{1}{243}\right)^6=\left(\dfrac{1}{3^5}\right)^6=\left(\dfrac{1}{3}\right)^{30}=\dfrac{1}{3^{30}}\)
Vì \(\dfrac{1}{3^{28}}>\dfrac{!}{3^{30}}\Rightarrow\left(\dfrac{1}{81}\right)^7>\left(\dfrac{1}{243}\right)^6\Rightarrow\) \(\left(\dfrac{1}{80}\right)^7>\left(\dfrac{1}{243}\right)^6\)
b, Ta có: \(\left(\dfrac{3}{8}\right)^5=\dfrac{3^5}{\left(2^3\right)^5}=\dfrac{243}{2^{15}}>\dfrac{243}{3^{15}}>\dfrac{125}{3^{15}}=\dfrac{5^3}{\left(3^5\right)^3}=\left(\dfrac{5}{243}\right)^3\)
\(\Rightarrow\left(\dfrac{3}{8}\right)^5>\left(\dfrac{5}{243}\right)^3\)
Giống nhau:
- Đều là các số tự nhiên
Khác nhau:
-số nguyên tố tự nhiên chỉ có hai ước là 1 và chính nó
-Hợp số là số tự nhiên có nhiều hơn hai ước
Tích của hai số nguyên tố là hợp số bởi ngoài ước là 1 ra nó còn có ước là hai số nguyên tố đó nữa.
A= \(\dfrac{\left(101+1\right)+\left(100+2\right)+...+\left(99+3\right)}{\left(101+1\right)-\left(100+2\right)+...+\left(99+3\right)-\left(98+3\right)}\)
= \(\dfrac{50.101}{50}\)
= 101
\(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(A=\dfrac{\left[\left(101-1\right):1+1\right].\left(101+1\right):2}{1.50+1}\)
\(A=\dfrac{5151}{51}=101\)
`A=3/4+8/9+.............+9999/10000`
`=1-1/4+1-1/9+,,,,,,,,,,+1-1/10000`
`=99-(1/4+1/9+.........+1/10000)<99-0=99`
`=>A<99`
Thanks