tinh:
M=1+1/2+1/22+1/23+1/24+......+1/298+1/299
luu y:/la phan
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đặt :
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{99}}\)
\(\Leftrightarrow2A=3+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{98}}\)
\(\Leftrightarrow2A-A=\left(3+\dfrac{1}{2}+....+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{2^{99}}\right)\)
\(\Leftrightarrow A=2-\dfrac{1}{2^{99}}\)
Vậy..
a: 12/y=4
nên y=12:4=3
b: Để 21/a;22/a-1;24/a+1 đều là số nguyên thì \(\left\{{}\begin{matrix}a\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\\a-1\in\left\{1;-1;2;-2;11;-11;22;-22\right\}\\a+1\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;8;-8;12;-12;24;-24\right\}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a\in\left\{1;-1;3;-3;7;-7;21;-21\right\}\\a\in\left\{2;0;3;-1;12;-10;23;-21\right\}\\a\in\left\{0;-2;1;-3;2;-4;3;-5;5;-7;7;-9;11;-13;23;-23\right\}\end{matrix}\right.\)
hay a=3
\(A=2^{100}-2^{99}+2^{98}-2^{97}+....-2^3+2^2-2+1\\ A=\left(2^{100}+2^{98}+...+2\right)-\left(2^{99}+2^{97}+...+1\right)\)
Gọi \(\left(2^{100}+2^{98}+...+2\right)\)là B
\(B=\left(2^{100}+2^{98}+...+2\right)\\ 2B=2^{102}+2^{100}+.....+2^2\\ 2B-B=\left(2^{102}+2^{100}+.....+2^2\right)-\left(2^{100}+2^{98}+...+2\right)\\ B=2^{102}-2\)
Gọi \(\left(2^{99}+2^{97}+...+1\right)\) là C
\(C=\left(2^{99}+2^{97}+...+1\right)\\ 2C=2^{101}+2^{99}+....+2\\ 2C-C=\left(2^{101}+2^{99}+9^{97}+...+2\right)-\left(2^{99}+9^{97}+...+1\right)\\ C=2^{101}-1\)
\(A=B+C\\ =>A=2^{102}-2+2^{101}-1\\ A=2^{101}\left(2+1\right)-3\\ A=2^{101}\cdot3-3\\ A=3\cdot\left(2^{101}-1\right)\)
\(\dfrac{1}{2}A=2^{99}-2^{98}+...-1+\dfrac{1}{2}\\ \Rightarrow A-\dfrac{1}{2}A=2^{100}-\dfrac{1}{2}\\ \Rightarrow A=2^{101}-1\)
a) \(A=1+2+2^2+...+2^{50}\)
\(\Rightarrow2A=2+2^2+...+2^{51}\)
\(\Rightarrow A=2A-A=2+2^2+...+2^{51}-1-2-2^2-...-2^{50}=2^{51}-1\)
b) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+...+3^{101}\)
\(\Rightarrow2B=3B-B=3+3^2+...+3^{101}-1-3-3^2-...-3^{100}=3^{101}-1\)
\(\Rightarrow B=\dfrac{3^{101}-1}{2}\)
c) \(C=5+5^2+...+5^{30}\)
\(\Rightarrow5C=5^2+5^3+...+5^{31}\)
\(\Rightarrow4C=5C-C=5^2+5^3+...+5^{31}-5-5^2-...-5^{30}=5^{31}-5\)
\(\Rightarrow C=\dfrac{5^{31}-5}{4}\)
d) \(D=2^{100}-2^{99}+2^{98}-...+2^2-2\)
\(\Rightarrow2D=2^{101}-2^{100}+2^{99}-...+2^3-2^2\)
\(\Rightarrow3D=2D+D=2^{101}-2^{100}+2^{99}-...+2^3-2^2+2^{100}-2^{99}+...+2^2-2=2^{101}-2\)
\(\Rightarrow D=\dfrac{2^{101}-2}{3}\)
Vẽ đồ thị hàm số:
1, y = 1/4x mũ 2
2, y = -1/4 x mũ 2
3, y = -2 x mũ 2
4, y = -1/2 x mũ 2
5, y = 3 x mũ 2
\(A=\dfrac{1}{\sqrt{25}+\sqrt{24}}+\dfrac{1}{\sqrt{24}+\sqrt{23}}+....+\dfrac{1}{\sqrt{2}+1}\)
\(A=\sqrt{25}-\sqrt{24}+\sqrt{24}-\sqrt{23}+......+\sqrt{2}-1=\sqrt{25}-1=4\)
1) \(\left(x+1\right)^2=x^2+2x+1\)
2) \(\left(2x+1\right)^2=4x^2+4x+1\)
3) \(\left(2x+y\right)^2=4x^2+4xy+y^2\)
4) \(\left(2x+3\right)^2=4x^2+12x+9\)
5) \(\left(3x+2y\right)^2=9x^2+12xy+4y^2\)
6) \(\left(2x^2+1\right)^2=4x^4+4x^2+1\)
7) \(\left(x^3+1\right)^2=x^6+2x^3+1\)
8) \(\left(x^2+y^3\right)^2=x^4+2x^2y^3+y^6\)
9) \(\left(x^2+2y^2\right)^2=x^4+4x^2y^2+4y^4\)
10) \(\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^2=\dfrac{1}{4}x^2+\dfrac{1}{3}xy+\dfrac{1}{9}y^2\)
ta có :
1/2=1/40+1/40+....+1/40 (20 số hạng)
1/21+1/22+1/23....+1/40(có 20 số hạng)
vì 1/21>1/40
1/22>1/40
..........
1/39>1/40
1/40=1/40
=>A<1/2
A<1 chịu
Ta có
\(\frac{1}{40}< \frac{1}{21}\\ \frac{1}{40}< \frac{1}{22}\\ ...\\ \frac{1}{40}< \frac{1}{39}\)
Mà số phần từ của A là 20
\(\Rightarrow\frac{1}{40}.20< A\Leftrightarrow\frac{1}{2}< A\)
Còn chứng minh bé hơn 1 thì tương tự bạn nhé!
\(2M=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}\)
Lấy 2M-M theo từng vế ta được:
\(2M-M=2+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}-...-\frac{1}{2^{98}}\)\(-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{99}}\)
\(M=2-\frac{1}{2^{99}}\)
Ko tính ra được kết quả cụ thể