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a, Ta có :
\(\left(\frac{1}{2}\right)^{50}=\left(\left(\frac{1}{2}\right)^5\right)^{10}=\left(\frac{1}{32}\right)^{10}\)
bạn so sánh nha :)
b,
T/c : \(99^{20}=\left(\left(99\right)^2\right)^{10}=9801^{10}\)
tiếp đây thì bạn tự làm nha có gì k hiểu ibx mk
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16 = 24
(\(\dfrac{1}{16}\))200 = \(\dfrac{1}{2^{4.200}}\) = \(\dfrac{1}{2^{800}}\)= (\(\dfrac{1}{2}\))800
So sánh với (\(\dfrac{1}{2}\))1000
Hai phân số cùng tử số, phân số nào có mẫu lớn hơn thì phân số đó nhỏ hơn
Suy ra: (\(\dfrac{1}{16}\))200 > (\(\dfrac{1}{2}\))1000
Ta có: \(\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2}\right)^{800}\)
mà \(\left(\dfrac{1}{2}\right)^{800}>\left(\dfrac{1}{2}\right)^{1000}\)
nên \(\left(\dfrac{1}{16}\right)^{200}< \left(\dfrac{1}{2}\right)^{1000}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(\dfrac{1}{2}\right)^{12}=\left(\dfrac{1}{8}\right)^3\\ \left(\dfrac{1}{3}\right)^9=\left(\dfrac{1}{27}\right)^3\\ Ta\text{ }có:\dfrac{1}{8}>\dfrac{1}{27}\\ Vậy:\left(\dfrac{1}{2}\right)^{12}>\left(\dfrac{1}{3}\right)^9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{2020^2}-1\right)\)
\(B=\left(\dfrac{1}{2^2}-\dfrac{2^2}{2^2}\right)\left(\dfrac{1}{3^2}-\dfrac{3^2}{3^2}\right)....\left(\dfrac{1}{2020^2}-\dfrac{2020^2}{2020^2}\right)\)
\(B=\left(\dfrac{1-2^2}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)...\left(\dfrac{1-2020^2}{2020^2}\right)\)
\(B=\dfrac{\left(1-2\right)\left(1+2\right)}{2^2}\cdot\dfrac{\left(1-3\right)\left(1+3\right)}{3^2}....\cdot\dfrac{\left(2020-1\right)\left(2020+1\right)}{2020^2}\)
\(B=\dfrac{-1\cdot3}{2^2}\cdot\dfrac{-2\cdot4}{3^2}\cdot\dfrac{-3\cdot5}{4^2}\cdot....\cdot\dfrac{-2019\cdot2021}{2020}\)
\(B=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-2019}{2\cdot3\cdot4\cdot....\cdot2020}\)
\(B=\dfrac{-1\cdot-1\cdot-1\cdot....\cdot-1}{1}\)
\(B=-1\) (2019 số -1)
Mà: \(-1< \dfrac{1}{2}\)
\(\Rightarrow B< \dfrac{1}{2}\)
\(\dfrac{1}{2^2}\); \(\dfrac{1}{3^2}\);...;\(\dfrac{1}{2020^2}\) < 1 ⇒ 0 > \(\dfrac{1}{2^2}\) - 1 > \(\dfrac{1}{3^2}\) - 1 >..> \(\dfrac{1}{2020^2}\) - 1
Xét dãy số 2; 3; 4;...; 2020 dãy số này có số số hạng là:
(2020 - 2):1 + 1 = 2019 (số hạng)
Vậy B là tích của 2019 số âm nên B < 0 ⇒ B < \(\dfrac{1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(K=\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\)
\(3K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}\)
\(3K-K=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{98}}-\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\)
\(2K=\)\(1-\frac{1}{3^{99}}\)
\(K=\frac{1-\frac{1}{3^{99}}}{2}\)
Có \(1-\frac{1}{3^{99}}\) < \(\frac{1}{2}\)
\(\Rightarrow K\) < \(\frac{1}{2}\)
Vậy \(\left(\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}\right)\) < \(\frac{1}{2}\)
(-1)99= -1
(-1)999= -1
Vì: -1= -1
=> (-1)99= (-1)999
\(\left(-1\right)^{99}=-1\)
\(\left(-1\right)^{999}=-1\)
\(\Rightarrow\left(-1\right)^{99}=\left(-1\right)^{999}\)