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Ta có: B= \(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\)
=> \(\frac{1}{2}B=\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{99}\right)\)
\(-\left(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+\left(\frac{1}{4}\right)^4+...+\left(\frac{1}{2}\right)^{99}+\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)\)
=> B - \(\frac{1}{2}B=\left(\frac{1}{2}+\left(\frac{1}{2}\right)^{99}\right)-\left(\left(\frac{1}{2}\right)^{100}+\left(\frac{1}{2}\right)^{100}\right)=\frac{1}{2}\)
=> B \(\times\left(1-\frac{1}{2}\right)=\frac{1}{2}\)
=> B = 1
Câu này chắc chắn đúng luôn
Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)
\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)
\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)
\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)
a)
\(5A=5+5^2+.....+5^{101}\)
\(\Rightarrow5A-A=\left(5+5^2+.....+5^{101}\right)-\left(1+5+.....+5^{100}\right)\)
\(\Rightarrow4A=5^{101}-1\)
\(\Rightarrow A=\frac{5^{101}-1}{4}\)
b)
\(2B=1+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2^2}+.....+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
Bài 1 lớp 7 không làm được thì chết đi
Bài 2:
4B=1.2.3.4+2.3.4.(5-1)+..........+(n-1).n.(n+1).[(n+2)-(n-2)]
4B=1.2.3.4+2.3.4.5-1.2.3.4+.......+(n-1).n.(n+1).(n+2)-(n-2).(n-1).n.(n+1)
4B=(n-1).n.(n+1).(n+2)
B=\(\frac{\left(n-1\right).n.\left(n+1\right).\left(n+2\right)}{4}\)
\(2.\)
\(a.\)
Ta có : \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)
\(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\)
\(\Rightarrow2^{332}< 3^{223}\)
\(b.\)
Ta có : \(90^{20}=\left(9^2\right)^{10}=81^{10}\)
Vì \(81^{10}< \) \(9999^{10}\)
\(\Rightarrow99^{20}< 9999^{10}\)
\(3.\)
\(a.\)
Ta có : \(\left(2x+1\right)^2=4\)
\(\Rightarrow2x+1=\pm\sqrt{4}=\pm2\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=2\\2x+1=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(b.\)
\(\left(3x-1\right)^3=27\)
\(\Rightarrow\left(3x-1\right)^3=3^3\)
\(\Rightarrow3x-1=3\)
\(\Rightarrow x=\dfrac{4}{3}\)
\(c.\)
\(\left(3x-1\right)^3=-\dfrac{8}{27}\)
\(\Rightarrow\left(3x-1\right)^3=\left(-\dfrac{2}{3}\right)^3\)
\(\Rightarrow3x-1=-\dfrac{2}{3}\)
\(\Rightarrow x=\dfrac{1}{9}\)
1 a) 2.16>2n>4 => 25>2n>22 => 5>n>2 => n=3;4
b) 9.27<3n<243 => 33<3n<35 => 3<n<5 => n=4
c) 125>5n+1>25 => 53>5n+1>52 =>3>n+1>2 => 3-1>n+1-1>2-1
=> 2>n>1 => không có giá trị nào của n để 2>n>1 khi n là số tự nhiên
2 a) 2332<2333 mà 2333=23.111=8111
3223>3222 mà 3222=32.111=9111
Vì 8111<9111 => 2333<3222 => 2332<3233
b) 9920=992.10=980110 mà 980110<999910 nên 9920<999910
3 a) (2x+1)2=4=22 => 2x+1=2 => x=\(\dfrac{1}{2}\)
b) (3x-1)3=27=33 => 3x-1=3 => x=\(\dfrac{4}{3}\)
c) (3x-1)3=-8/27=(-2/3)3 => 3x-1=-2/3 => x=\(\dfrac{1}{9}\)
Trả lời: 0.5