theo minh thi

neu a<b thi ta co a(b+n) va b(a+n)

       ab+an và ab + bn

vi a<b nen a(b+n)<b(a+n) suy ra a/b < a+n/b+n

neu a>b thi ta co a(b+n) va b(a+n)

      ab+an va ab+bn

vì a>b nen a(b+n)>b(a+n) suy ra a/b>a+n/b+n

neu a=b thi a(b+n) và b(a+n)

       ab+an và ab+ bn

vì a=b nên a(b+n) = b(a+n) suy ra a/b=a+n/b+n

a bé hơn b

a+n<b+n
 

mik ko biết làm nhưng bạn có thể vào câu hỏi tương tự

Ta có : \(\frac{a}{b}< \frac{a+n}{b+n}\Leftrightarrow a(b+n)< b(a+n)\)

\(\Leftrightarrow ab+an< ab+bn\Leftrightarrow a< b\)vì n > 0

Như vậy : \(\frac{a}{b}< \frac{a+n}{b+n}\Leftrightarrow a< b\)

Ta lại có : \(\frac{a}{b}>\frac{a+n}{b+n}\Leftrightarrow a(b+n)>b(a+n)\)

\(\Leftrightarrow ab+an>ab+bn\Leftrightarrow an>bn\Leftrightarrow a>b\)

Như vậy : \(\frac{a}{b}>\frac{a+n}{b+n}\Leftrightarrow a>b\)

Ta có:a/b=a.(b+n)

                =a.b+a.n/b.(b+n)

a+n/b+n=(a+n).b/(b+n).b

             =a.b+b.n/b.(b+n)

-->a/b<a+n/b+n

Lời giải:

Xét $\frac{a}{b}-\frac{a+n}{b+n}=\frac{a(b+n)-b(a+n)}{b(b+n)}=\frac{n(a-b)}{b(b+n)}$
Nếu $a>b$ thì ${a}{b}-\frac{a+n}{b+n}=\frac{n(a-b)}{b(b+n)}>0$

$\Rightarrow {a}{b}>\frac{a+n}{b+n}$

Nếu $a=b$ thì ${a}{b}-\frac{a+n}{b+n}=\frac{n(a-b)}{b(b+n)}=0$

$\Rightarrow {a}{b}=\frac{a+n}{b+n}$

Nếu $a<b$ thì ${a}{b}-\frac{a+n}{b+n}=\frac{n(a-b)}{b(b+n)}<0$

$\Rightarrow {a}{b}<\frac{a+n}{b+n}$

\(a,\dfrac{a}{b}>1\Leftrightarrow a>1\cdot b=b\\ \dfrac{a}{b}< 1\Leftrightarrow a< 1\cdot b=b\\ b,\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{ab+a}{b^2+b}\\ \dfrac{a+1}{b+1}=\dfrac{b\left(a+1\right)}{b\left(b+1\right)}=\dfrac{ab+b}{b^2+b}\\ \forall a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+1}{b+1}\\ \forall a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+1}{b+1}\\ \forall a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+1}{b+1}\)

\(c,\forall a>b\Leftrightarrow\dfrac{a}{b}-1=\dfrac{a-b}{b}>\dfrac{a-b}{b+n}\left(b< b+n;a-b>0\right)=\dfrac{a+n}{b+n}-1\\ \Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a< b\Leftrightarrow1-\dfrac{a}{b}=\dfrac{b-a}{b}>\dfrac{b-a}{b+n}\left(b< b+n;b-a>0\right)=1-\dfrac{a+n}{b+n}\\ \Leftrightarrow1-\dfrac{a}{b}>1-\dfrac{a+n}{b+n}\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\\ \forall a=b\Leftrightarrow\dfrac{a+n}{b+n}=\dfrac{a}{b}\left(=1\right)\)