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24 tháng 4 2023
A=1/4(1/1+1/2^2+...+1/50^2)
=>A=1/4+1/4*(1/2^2+...+1/50^2)
=>A<1/4+1/4*(1-1/2+1/2-1/3+...+1/49-1/50)
=>A<1/4+1/4*49/50=99/200<1/2
TB
12 tháng 3 2019
C/m nó nhỏ hơn 3/4 hả bạn ?
Có \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{100}< \frac{3}{4}\)
XO
8 tháng 7 2021
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)(ĐPCM)
11 tháng 2 2016
2x(3y-2)+(3y-2) = (2x+1)(3y-2) = -55.Lập bảng :
| 2x+1 | -55 | -11 | -5 | -1 | 1 | 5 | 11 | 55 |
| 3y-2 | 1 | 5 | 11 | 55 | -55 | -11 | -5 | -1 |
| 2x | -56 | -12 | -6 | -2 | 0 | 4 | 10 | 54 |
| 3y | 3 | 7 | 13 | 57 | -53 | -9 | -3 | 1 |
| x | -28 | -6 | -3 | -1 | 0 | 2 | 5 | 27 |
| y | 1 | 19 | -3 | -1 |
Vậy (x;y) = (-28;1);(-1;19);(2;-3);(5;-1)
12 tháng 12 2019
a) Ta có : A=2+22+23+...+210
=(2+22)+(23+24)+...+(29+210)
=2(1+2)+23(1+2)+...+29(1+2)
=2.3+23.3+...+29.3
Vì 3\(⋮\)3 nên 2.3+23.3+...+29.3\(⋮\)3
hay A\(⋮\)3
Vậy A\(⋮\)3.
NT
1
PM
0
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)
\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)
Đặt \(B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\)
\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)
Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
=>\(B=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 2-\dfrac{1}{50}\)
=>\(A=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{2^2}\left(2-\dfrac{1}{50}\right)=\dfrac{1}{2}-\dfrac{1}{200}< \dfrac{1}{2}\)
\(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{6}\right)^2+...+\left(\dfrac{1}{100}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}.\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}.\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{2}.\dfrac{1}{50}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^2.\left[1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2\right]\)
Ta có:
\(\left(\dfrac{1}{2}\right)^2=\dfrac{1}{2.2}< \dfrac{1}{2.1}=\dfrac{2-1}{2.1}=\dfrac{2}{2.1}-\dfrac{1}{2.1}=1-\dfrac{1}{2}\)
\(\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3.3}< \dfrac{1}{3.2}=\dfrac{3-2}{3.2}=\dfrac{3}{3.2}-\dfrac{2}{3.2}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\left(\dfrac{1}{50}\right)^2=\dfrac{1}{50.50}< \dfrac{1}{50.49}=\dfrac{50-49}{50.49}=\dfrac{50}{50.49}-\dfrac{49}{50.49}=\dfrac{1}{49}-\dfrac{1}{50}\)
Khi đó
\(1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2< 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)
\(=\left(\dfrac{1}{2}\right)^2.\left[1+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{3}\right)^2+...+\left(\dfrac{1}{50}\right)^2\right]< \dfrac{1}{4}.2=\dfrac{1}{2}\)
Vậy \(\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{6}\right)^2+...+\left(\dfrac{1}{100}\right)^2< \dfrac{1}{2}\left(đpcm\right)\)
Tick cho mk nha :>>