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

\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{4031}{2015^2.2016^2}=\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{2^2}-.....-\frac{1}{2016^2}=1-\frac{1}{2016^2}\)

\(\frac{1}{2016^2}>0\Rightarrow A< 1\left(ĐPCM\right)\)

bạn chờ xíu mk lm câu sau nha

Bạn chờ xíu mk lm cho xong 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: \(x>y>z\)

\(\Rightarrow\frac{y}{y+z}+\frac{z}{z+x}< \frac{x+y}{y+z}=1;\frac{x}{x+y}< 1\Rightarrow P< 1+1=2\Rightarrow1< P< 2\left(ĐPCM\right)\)

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\left(dpcm\right)\) 

CS AI XEM S** KO

\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+...+\dfrac{19}{81.100}\)\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{81}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}< 1\)