Question

[Quỳnh Anh]

a) Cho phân số \(\frac{a}{b}\) ( a, b \(\in\)N , b \(\ne\)0 ) 

Giả sử  \(\frac{a}{b}\)> 1  và m \(\in\) N , m \(\ne\) 0 . Chứng tỏ rằng :

\(\frac{a}{b}\) > \(\frac{a+m}{b+m}\)

b) Áp dụng kết quả ở câu a) để so sánh \(\frac{237}{142}\) và \(\frac{246}{151}\)

Answer 1

[Đinh Đức Hùng]

a ) Nếu \(\frac{a}{b}>\frac{a+m}{b+m}\)

\(\Leftrightarrow a\left(b+m\right)>b\left(a+m\right)\)

\(\Leftrightarrow ab+am>ab+bm\)

\(\Leftrightarrow am>bm\)

\(\Rightarrow a>b\)

\(\Rightarrow\frac{a}{b}>1\)

Vậy \(\frac{a}{b}>1\) thì \(\frac{a}{b}>\frac{a+m}{b+m}\)

b ) Vì 237 > 142 => \(\frac{237}{142}>\frac{237+9}{142+9}=\frac{246}{151}\)

Answer 2

[Trần Quỳnh Mai]

Xét hiệu :

\(\frac{a}{b}-\frac{a+m}{b+m}\)

\(=\frac{a\left(b+m\right)}{b\left(b+m\right)}-\frac{\left(a+m\right)b}{\left(b+m\right)b}\)

\(=\frac{a.b+a.m}{b\left(b+m\right)}-\frac{a.b+b.m}{b\left(b+m\right)}\)

\(=\frac{a.b+a.m-a.b+b.m}{b\left(b+m\right)}\)

\(=\frac{m\left(a-b\right)}{b\left(b+m\right)}\)

Vì \(\frac{a}{b}>1,b\in\)N* \(\Rightarrow a>b\Rightarrow a-b>0,m\in\)N*

\(\Rightarrow m\left(a-b\right)>0\); Vì : \(b,m\in\)N* \(\Rightarrow b\left(b+m\right)>0\)

\(\Rightarrow\frac{m\left(a-b\right)}{b\left(b+m\right)}>0\) hay : \(\frac{a}{b}-\frac{a+m}{b+m}>0\Rightarrow\frac{a}{b}>\frac{a+m}{b+m}\)

Vậy \(\frac{a}{b}>1,m\in\)N* thì \(\frac{a}{b}>\frac{a+m}{b+m}\)

b, Tự làm