Cho A=1/2!+2/3!+3/4!+...+2015/2016!
Chung minh :A<1
Cac ban oi giup minh di
Thank
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hinh nhu trong sach phat trien lop 6 co thi phai,lau roi quen
a. \(A=\left[\frac{1}{3}+\frac{3}{x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(9-x^2\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x.\left(x-3\right)}{3.x.\left(x-3\right)}+\frac{3.3}{x\left(x-3\right).3}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x^2-3x+9}{3x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{\left(3-x\right).3}{\left(x+3\right).\left(3-x\right).3}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}:\left[\frac{x^2+9-3x}{3.\left(3-x\right)\left(3+x\right)}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}.\frac{3.\left(3-x\right)\left(3+x\right)}{x^2-3x+9}\)
\(=\frac{-\left(x-3\right)\left(3+x\right)}{x-3}=-\left(3+x\right)\)
b. Để A < -1 thì:
-(3+x) < -1
=> -3 - x < -1
=> x < -3 - (-1) = -2
Vậy x < -2 thì A < -1.
\(\frac{x+1}{2017}+\frac{x+2}{2016}=\frac{x+3}{2015}+\frac{x+4}{2014}\)
\(\Leftrightarrow\frac{x+1}{2017}+1+\frac{x+2}{2016}+1=\frac{x+3}{2015}+1+\frac{x+4}{2014}+1\)
\(\Leftrightarrow\frac{x+2018}{2017}+\frac{x+2018}{2016}-\frac{x+2018}{2015}-\frac{x+2018}{2014}=0\)
\(\Leftrightarrow\left(x+2018\right)\left(\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}\ne0\right)=0\Leftrightarrow x=-2018\)
Ta có : \(\dfrac{1}{2^2}\)<\(\dfrac{1}{1.2}\); \(\dfrac{1}{3^2}\)<\(\dfrac{1}{2.3}\);.....;\(\dfrac{1}{2016^2}\)<\(\dfrac{1}{2015.2016}\)
⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\)< \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{2015.2016}\)
⇒ A = \(\dfrac{1}{2^2}\)+\(\dfrac{1}{3^2}\)+...+\(\dfrac{1}{2016^2}\) < 1 - \(\dfrac{1}{2016}\)= \(\dfrac{2015}{2016}\) (ĐCPCM)
Vì các số hạng trong tổng trên đều < 1
Nên => A<1
1.2!+2/3!+...đều là tổng các phân số có tử là 1. 1/2!=1/2
2/3!=1/3; 3/4!=1/8 .... nên tổng A bé hơn 1