Question

cho a,b thuộc Z, b>0 .so sánh hai số hữu tỉ a/b và a+2001/b+2001

Answer 1

a/b < a+2001/b+2001

Answer 2

Ta có: \(\frac{a}{b}=\frac{a.\left(b+2001\right)}{b.\left(b+2001\right)}=\frac{ab+2001a}{b^2+2001b}\) 

         \(\frac{a+2001}{b+2001}=\frac{b.\left(a+2001\right)}{b.\left(b+2001\right)}=\frac{ab+2001b}{b^2+2001b}\)

*TH1: a=b

=>\(\frac{a}{b}=\frac{a+2001}{b+2001}=1\)

*TH2: a<b

=>ab+2001a<ab+2001b

=>\(\frac{ab+2001a}{b^2+2001b}< \frac{ab+2001b}{b^2+2001b}\)

=>\(\frac{a}{b}< \frac{a+2001}{b+2001}\)

TH3:a>b

=>ab+2001a>ab+2001b

=>\(\frac{ab+2001a}{b^2+2001b}>\frac{ab+2001b}{b^2+2001b}\)

=>\(\frac{a}{b}>\frac{a+2001}{b+2001}\)

Answer 3

Ta có: a(b + 2001) = ab + 2001a

         : b(a + 2001) = ab + 2001b

-Trường hợp 1: Nếu a > b \(\Rightarrow\)2001a > 2001b

 \(\Rightarrow\)ab + 2001a > ab + 2001b \(\Rightarrow\)\(\frac{a}{b}>\frac{a+2001}{b+2001}\)

-Trường hợp 2: Nếu a < b, tương tự ta có: \(\frac{a}{b}< \frac{a+2001}{b+2001}\)

-Trường hợp 3: Nếu a = b \(\Rightarrow\) \(\frac{a}{b}=\frac{a+2001}{b+2001}\)

Chúc bạn học tốt ^^!