Question

[nguyễn thùy trang]

Bài 3 

a) Với mọi số tự nhiên a,b,c khác 0,chứng minh:

* Nếu a/b < 1 thì a/b < a+c/b+c

* Nếu a/b > 1 thì a/b > a+c/b+c

b) So sánh A và B biết : A = 2008 ^ 2008 + 1 / 2008 ^ 2009 + 1

                                      B = 2008 ^ 2007 + 1 / 2008 ^ 2008 + 1

giúp mình với mọi người ơi mình cần gấp

Answer 1

[HT.Phong (9A5)]

a) \(\dfrac{a}{b}< 1=>a< b\) 

\(\dfrac{a}{b}-\dfrac{a+c}{b+c}=\dfrac{a\left(b+c\right)-b\left(a+c\right)}{b\left(b+c\right)}=\dfrac{ab+ac-ab-bc}{b\left(b+c\right)}=\dfrac{c\left(a-b\right)}{b\left(b+c\right)}\)

Mà: \(a< b=>a-b< 0=>\dfrac{c\left(a-b\right)}{b\left(b+c\right)}< 0\)

\(=>\dfrac{a}{b}-\dfrac{a+c}{b+c}< 0=>\dfrac{a}{b}< \dfrac{a+c}{b+c}\)

______________________________

\(\dfrac{a}{b}>1=>a>b\)

\(\dfrac{a}{b}-\dfrac{a+c}{b+c}=\dfrac{a\left(b+c\right)-b\left(a+c\right)}{b\left(b+c\right)}=\dfrac{ab+ac-ab-bc}{b\left(b+c\right)}=\dfrac{c\left(a-b\right)}{b\left(b+c\right)}\)

Mà: \(a>b=>a-b>0=>\dfrac{c\left(a-b\right)}{b\left(b+c\right)}>0\)

\(=>\dfrac{a}{b}-\dfrac{a+c}{b+c}>0=>\dfrac{a}{b}>\dfrac{a+c}{b+c}\)

b) Ta có: 

\(2008^{2008}+1< 2008^{2009}+1=>A=\dfrac{2008^{2008}+1}{2008^{2009}+1}< 1\) 

Áp dụng CT ở câu a \(\dfrac{a}{b}< 1=>\dfrac{a}{b}< \dfrac{a+c}{b+c}\) 

\(=>A< \dfrac{2008^{2008}+1+2007}{2008^{2009}+1+2007}=\dfrac{2008^{2008}+2008}{2008^{2009}+2008}\\ =>A< \dfrac{2008\left(2008^{2007}+1\right)}{2008\left(2008^{2008}+1\right)}\\ =>A< \dfrac{2008^{2007}+1}{2008^{2008}+1}\\ =>A< B\)

Answer 2

[Dương Minh Uy]

a)Nếu \(\dfrac{a}{b}< 1\) thì  \(\dfrac{a+c}{b+c}=\dfrac{\dfrac{a}{b}+\dfrac{c}{b}}{\dfrac{b}{b}+\dfrac{c}{b}}>\dfrac{1+\dfrac{c}{b}}{1+\dfrac{c}{b}}=1>\dfrac{a}{b}\)

b) Nếu \(\dfrac{a}{b}>1\) thì  \(\dfrac{a+c}{b+c}=\dfrac{\dfrac{a}{b}+\dfrac{c}{b}}{\dfrac{b}{b}+\dfrac{c}{b}}>\dfrac{1+\dfrac{c}{b}}{1+\dfrac{c}{b}}=1>\dfrac{a}{b}\)

c)Áp dụng

 \(A=\dfrac{2008^{2008}+1}{2008^{2009}+1}< \dfrac{2008^{2008}+1+2007}{2008^{2009}+1+2007}=\dfrac{2008\left(2^{2007}+1\right)}{2008\left(2^{2008}+1\right)}=\dfrac{2008^{2007}+1}{2008^{2008}+1}=B\)