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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Để so sánh a/b và a+2016/b+2016 ta xét hiệu.
\(H=\frac{a}{b}-\frac{a+2016}{b+2016}=\frac{ab+2016a-ab-2016b}{b\left(b+2016\right)}=\frac{2016\left(a-b\right)}{b\left(b+2016\right)}.\)
- Do b dương, nên H dương khi a>b =>\(\frac{a}{b}>\frac{a+2016}{b+2016}\)
- H âm khi a<b => \(\frac{a}{b}< \frac{a+2016}{b+2016}\)
- H = 0 khi a=b => \(\frac{a}{b}=\frac{a+2016}{b+2016}\)
Mở rộng bài toán ta được: \(\forall a;b\in R;b>0;m>0\) thì
- \(\frac{a}{b}>\frac{a+m}{b+m}\)khi \(a>b\);
- \(\frac{a}{b}< \frac{a+m}{b+m}\)khi \(a< b\);
- \(\frac{a}{b}=\frac{a+m}{b+m}\)khi \(a=b\);
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét hiệu:
\(H=\frac{a}{b}-\frac{a+2016}{b+2016}=\frac{a\cdot\left(b+2016\right)-\left(a+2016\right)\cdot b}{b\left(b+2016\right)}=\frac{2016\cdot\left(a-b\right)}{b\left(b+2016\right)}.\)
- Nếu b<-2016 và a>b thì H>0; a<b thì H<0
- -2016<b<0 và a>b thì H<0; a<b thì H>0
- Nếu b>0 và a>b thì H>0; a<b thì H<0
tùy H>0 hay H<0 mà ta biết được kq của sự so sánh.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b+2016}< \frac{a}{b}\) và \(\frac{2016}{b+2016}< \frac{a}{b}\)
=> \(\frac{a}{b+2016}+\frac{2016}{b+2016}< \frac{a}{b}\)
hay \(\frac{a+2016}{b+2016}< \frac{a}{b}\)
n
nếu a>b hay a/b > 1 ta có 2016a > 2016b
=> 2016a + ab > 2016b + ab
=> a ( 2016 + b) > b ( 2016 + a )
=> a/b > a+2016/b+2016
tương tự với 2 trường hợp
nếu a < b thì a/b < a+2016/b+2016
nếu a = b thì a/b = a+2016/b+2016
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\frac{a}{b}-1=\frac{a}{b}-\frac{b}{b}=\frac{a-b}{b}\)
\(\frac{a+2016}{b+2016}-1=\frac{a+2016}{b+2016}-\frac{b+2016}{b+2016}=\frac{a+2016-b-2016}{b+2016}=\frac{a-b}{b+2016}\)
So sánh nứa là ra ok bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{a}{b}\)= \(\frac{a\left(b+2016\right)}{b\left(b+2016\right)}\)=\(\frac{ab+2016a}{b\left(b+2016\right)}\)
\(\frac{a+2016}{b+2016}\)=\(\frac{\left(a+2016\right)b}{\left(b+2016\right)b}\)=\(\frac{ab+2016b}{b\left(b+2016\right)}\)
Vì b > 0 nên mẫu số của hai phân số trên dương. Ta so sánh tử số.
* Nếu a < b => ab+2016a < ab+2016b
=> \(\frac{a}{b}\)<\(\frac{a+2016}{b+2016}\)
* Nếu a = b => ab+2016a = ab+2016b
=> \(\frac{a}{b}\)=\(\frac{a+2016}{b+2016}\)
* Nếu a > b => ab+2016a > ab+2016b
=> \(\frac{a}{b}\)>\(\frac{a+2016}{b+2016}\)
Giả sử a/b = 1/3 còn phân số kia là 2017/2019
quy đồng 1/3 = 2017/6051
Vì 6051>2019 nên 2017/2019 > 2017/6051 và 2017/2019>1/3
Vậy \(\frac{a}{b}< \frac{a+2016}{b+2016}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
+\(\frac{a}{b}=1\Leftrightarrow a=b\Leftrightarrow\frac{a}{b}=\frac{a+2016}{b+2016}\)
+\(\frac{a}{b}>1\Leftrightarrow a>b\Leftrightarrow\frac{a}{b}-1=\frac{a-b}{b}>\frac{a-b}{b+2016}=\frac{a+2016}{b+2016}-1\)=> \(\frac{a}{b}>\frac{a+2016}{b+2016}\)
+\(\frac{a}{b}< 1\Leftrightarrow a< b\Leftrightarrow1-\frac{a}{b}=\frac{b-a}{b}>\frac{b-a}{b+2016}=1-\frac{a+2016}{b+2016}\)=>\(\frac{a}{b}< \frac{a+2016}{b+2016}\)
Xét 3 trường hợp
TH1 \(a=b\)\(\Rightarrow\frac{a}{b}=1=\frac{a+2016}{b+2016}\)
TH2 \(a>b\Rightarrow\frac{a}{b}>1\)\(\Rightarrow2016a>2016b\)
Ta có: \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016a}{b\left(b+2016\right)}\)
\(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016b}{b\left(b+2016\right)}\)
Ta có: 2016a>2016b => ab+2016a>ab+2016b hay a/b>a+2016/b+20116
TH3
\(a< b\Rightarrow\frac{a}{b}< 1\)\(\Rightarrow2016a< 2016b\)
Ta có: \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016a}{b\left(b+2016\right)}\)
\(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}=\frac{ab+2016b}{b\left(b+2016\right)}\)
Ta có: 2016a<2016b => ab+2016a<ab+2016b hay a/b<a+2016/b+20116
a b 0 ta co 1 2 3 4 5+2 3 4 5 6