CHỨNG MINH RẰNG : \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4031}{2015^2.2016^2}< 1\)
chứng minh \(A=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{4031}{2015^2.2016^2}< 1\)
CHỨNG MINH RẰNG
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......+\frac{4031}{2015^2.2016^2}< 1\)
c/minh: A=3/1^2.2^2+5/2^2.3^2+7/3^2.4^2+.......+4031/2015^2.2016^2<1
A =2^2-1^2/1^2.2^2 + 3^2-2^2/2^2.3^2 + ..... + 2016^2-2015^2/2015^2.2016^2
= 1/1^2-1/2^2+1/2^2-1/3^2+.....+1/2015^2-1/2016^2
= 1-1/2016^2 < 1
=> ĐPCM
k mk nha
Mk hơi bối rối,bn dùng cái gõ phương trình trên thanh công cụ được ko.
(2x)^2 - 25=0
-> (2x)^2 = 0+ 25 = 25
-> (2x) = 5
Vậy x = 5:2 = 2.5
Chứng minh rằng: 3/1^2.2^2 + 5/2^2.3^2 + 7/3^2.4^2 + ... + 4031/2015^2.2016^2 < 1
Ta có: \(\frac{3}{1^2.2^2}=\frac{1}{1^2}-\frac{1}{2^2}\); \(\frac{5}{2^2.3^2}=\frac{1}{2^2}-\frac{1}{3^2}\); \(\frac{7}{3^2.4^2}=\frac{1}{3^2}-\frac{1}{4^2}\);....; \(\frac{4031}{2015^2.2016^2}=\frac{1}{2015^2}-\frac{1}{2016^2}\)
=> \(A=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{2015^2}-\frac{1}{2016^2}\)
=> \(A=1-\frac{1}{2016^2}< 1\)
=> A < 1
Câu 1
Chứng minh rằng: A=\(\frac{3}{1^2.2^2}\) + \(\frac{5}{2^2.3^2}\) + \(\frac{7}{3^2.4^2}\) + ... + \(\frac{4031}{2015^2.2016^2}\) < 1
Câu 2
Cho biểu thức P = \(\frac{x}{x+y}\) + \(\frac{y}{y+z}\) + \(\frac{z}{z+x}\) với x, y, z là các số nguyên dương. Chứng minh 1 < P < 2.
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{4031}{2015^2.2016^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-.....-\frac{1}{2016^2}=1-\frac{1}{2016^2}\)
\(\frac{1}{2016^2}>0\Rightarrow A< 1\left(ĐPCM\right)\)
bạn chờ xíu mk lm câu sau nha
\(Taco:\)
\(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x};x,y,z\inℕ^∗\)
\(\frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}>\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}=1\)
\(\Rightarrow P>1\)
Giả sử: \(x>y>z\)
\(\Rightarrow\frac{y}{y+z}+\frac{z}{z+x}< \frac{x+y}{y+z}=1;\frac{x}{x+y}< 1\Rightarrow P< 1+1=2\Rightarrow1< P< 2\left(ĐPCM\right)\)
Chứng minh rằng : \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\)
Xét số bất kì a. Ta sẽ chứng mỉnh (a + 1)2 - a2 = 2a + 1.
Thật vậy, ta có (a + 1)2 - a2 = a(a + 1) + (a + 1) - a2 = (a2 + a) + (a + 1) = 2a + 1 (đpcm).
Áp dụng ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=\frac{1}{1^2}-\frac{1}{10^2}< 1\left(đpcm\right)\)
Chứng minh rằng: \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+......+\frac{19}{9^2.10^2}< 1\)
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
= \(\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
= \(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
= \(1-\frac{1}{10^2}\)< 1
Vậy
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\) <1
chứng minh rằng : \(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}< 1\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+.....+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+.....+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{10^2-9^2}{9^2.10^2}\)
\(=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{9^2}-\frac{1}{10^2}\)
\(=1-\frac{1}{10^2}< 1\)
Chứng minh rằng: A=\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}<1\)