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1, Thấy : \(\frac{1}{5}< \frac{2}{2.4}\)
\(\frac{1}{13}< \frac{2}{4.6}\)
.....
\(\frac{1}{n^2+\left(n+1\right)^2}< \frac{2}{2n\left(2n+1\right)}\)
Cộng từng vế có :
\(\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2n\left(2n+2\right)}\)
\(\frac{1}{5}+\frac{1}{13}+...+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}-\frac{1}{4}+....+\frac{1}{2n}-\frac{1}{2n+2}\)
\(\frac{1}{5}+\frac{1}{13}+..+\frac{1}{n^2+\left(n+1\right)^2}< \frac{1}{2}-\frac{1}{2n+2}\)
Mà \(\frac{1}{2}-\frac{1}{2n+2}< \frac{1}{2}\)=> Tổng trên < 1/2
2,M = \(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
=> M \(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+...+\frac{1}{\left(n-1\right)^2}-\frac{1}{n^2}+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)
\(M=1-\frac{1}{\left(n+1\right)^2}=\frac{\left(n+1\right)^2-1}{\left(n+1\right)^2}=\frac{n^2+2n+1-1}{\left(n+1\right)^2}=\frac{n^2+2n}{\left(n+1\right)^2}\)
Đến đây tắc r tự nghĩ tiếp >:
`2x-15=-25`
`2x=-10`
`x=-5`
___________
`3/5<x/10<4/5`
`3/5=(3xx10)/(5xx10)=30/50`
`x/10=(5x)/(10xx5)=(5x)/50`
`4/5=(4xx10)/(5xx10)=40/50`
`=>30/50<(5x)/50<40/50`
`=>30<5x<40`
`=>x=7`
a: Ta có: \(3n+2⋮n-1\)
\(\Leftrightarrow n-1\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{2;0;6;-4\right\}\)
a. Tìm x thuộc N sao cho : 2x + 1 thuộc Ư ( 2x + 10)
(2x + 10) ⋮ (2x + 1)
Ta có (2x + 10) = (2x + 1 + 9)
Mà (2x + 10) ⋮ (2x + 1)
Nên 9 ⋮ (2x + 1)
Do đó ta có (2x + 1) ∈ Ư (9) = {-1; 1; -3; 3; -9; 9}
2x + 1 | -1 | 1 | -3 | 3 | -9 | 9 |
2x | -2 | 0 | -4 | 2 | -10 | 8 |
x | -1 | 0 | -2 | 1 | -5 | 4 |
Vậy x = {-1; 0; -2; 1; -5; 4}
b. A = 3 - 5 + 13 - 15 + 23 - 25 + ....... + 93 - 95 + 2020
A = (3 - 5) + (13 - 15) + (23 - 25) +.......+ (93 - 95) + 2020
A = (-2) + (-2) + (-2) + ......... + (-2) + 2020
Có 10 số (-2)
A = (-2) . 10 + 2020
A = (-20) + 2020
A = 2000
c. 2( x + 1) - x - 2 = (-5) - 3
2x + 2 - 1x - 2 = (-8)
2x - 1x + 2 - 2 = (-8)
2x - 1x + 0 = (-8)
2x - 1x = (-8)
(2 - 1)x = (-8)
1 . x = (-8) : 1
x = (-8)
a: \(\Leftrightarrow n+1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(n\in\left\{0;-2;1;-3;2;-4;5;-7\right\}\)
b: \(\Leftrightarrow n-1\in\left\{1;-1;7;-7\right\}\)
hay \(n\in\left\{2;0;8;-6\right\}\)
a, \(n^2+5=n^2+n-n-1+6=n\left(n+1\right)-\left(n+1\right)+6\)
\(\Rightarrow n+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
n + 1 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
n | 0 | -2 | 1 | -3 | 2 | -4 | 5 | -7 |
b, tương tự
a 25 + 8 + 12 + 25 = 70 b 87 + [ - 750 ] + 2018 - 13 + 750 = 2842 c [ 45 + 12 - 13 ] - [ 45 - 13 + 12 ] = 90 minh tinh gom bn ah