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Bài 1:
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(\Rightarrow2A=2+\frac{3}{2^2}+\frac{4}{2^3}+....+\frac{100}{2^{99}}\)
\(\Rightarrow2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\right)\)
\(\Rightarrow A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(\Rightarrow A=1+\frac{3}{2^2}+\left(\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
Bài 2:
Giải:
Ta có: \(2n-3⋮n+1\)
\(\Rightarrow\left(2n+2\right)-5⋮n+1\)
\(\Rightarrow2\left(n+1\right)-5⋮n+1\)
\(\Rightarrow5⋮n+1\)
\(\Rightarrow n+1\in\left\{1;-1;5;-5\right\}\)
\(\Rightarrow n\in\left\{0;-2;4;-6\right\}\)
Vậy ...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Câu hỏi của Nguyễn Khánh Ly - Toán lớp 7 - Học toán với OnlineMath
b) 2n - 3 = 2n + 2 - 5 chia hết cho n + 1
<=> 5 chia hết cho n + 1
<=> n + 1 thuộc Ư(5) = {1;5}
<=> n thuộc {0;4}
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a: \(3x-\left|2x+1\right|=2\)
\(\Leftrightarrow\left|2x+1\right|=3x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2\right)^2-\left(2x+1\right)^2=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(3x-2-2x-1\right)\left(3x-2+2x+1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)\left(5x-1\right)=0\\x>=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow x=3\)
e: Ta có: \(2n-3⋮n+1\)
\(\Leftrightarrow2n+2-5⋮n+1\)
\(\Leftrightarrow n+1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{0;-2;4;-6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(2A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
\(2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\right)\)
\(A=1+\frac{3}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(A=\frac{7}{4}-\frac{100}{2^{100}}+\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\right)\)
Đặt \(B=\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
\(2B=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}\)
\(2B-B=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\right)\)
\(B=\frac{1}{2^2}-\frac{1}{2^{99}}\)
\(\Rightarrow\)\(A=\frac{7}{4}-\frac{100}{2^{100}}+B=\frac{7}{4}-\frac{100}{2^{100}}+\frac{1}{2^2}-\frac{1}{2^{99}}=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}=\frac{2^{101}-102}{2^{100}}\)
Vậy \(A=\frac{2^{101}-102}{2^{100}}\)
Chúc bạn học tốt ~
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1. A = 75(42004 + 42003 +...+ 42 + 4 + 1) + 25
A = 25 . [3 . (42004 + 42003 +...+ 42 + 4 + 1) + 1]
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 3 + 1)
A = 25 . (3 . 42004 + 3 . 42003 +...+ 3 . 42 + 3 . 4 + 4)
A = 25 . 4 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1)
A =100 . (3 . 42003 + 3 . 42002 +...+ 3 . 4 + 3 + 1) \(⋮\) 100
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Bạn ghi nhỏ lại nhé. Hơn nũa bạn nên tách riêng từng câu hỏi, làm vầy nhiều lắm
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3n+2 - 2n+2 +3n - 2n = 3n . 32 - 2n. 22 +3n -2n
= 3n(32+1) - (2n.22 +2n)
=3n . 10 - 2n .5
=3n.10 - 2n-1 .2 .5
= 3n.10 - 2n-1 .10
= 10(3n - 2n-1)
vì 10 chia hết cho 10 nên 10(3n-2n-1) chia hết cho 10
=> 3n+2 - 2n+2 +3n -2n chia hết cho 10
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