Ta có:\(\frac{a}{b}=\frac{a.\left(b+2012\right)}{b.\left(b+2012\right)}=\frac{ab+a.2012}{b.\left(b+2012\right)}\)

\(\frac{a+2012}{b+2012}=\frac{b.\left(a+2012\right)}{b.\left(b+2012\right)}=\frac{ab+b.2012}{b.\left(b+2012\right)}\)

Vì a<0<b=>a<b=>a.2012<b.2012

=>\(\frac{ab+a.2012}{b.\left(b+2012\right)}

ko hiểu đề

để so sánh a/b và a+2012/b+2012

Ta xét tích:a(b+2012) và b(a+2012)

Vì b>0 =>b+2012>0

*a>b <=>2012a>2012b

<=>a(b+2012)>b(a+2012)

<=>a/b>a+2012/b+2012

*a=b<=>2012a=2012b

<=>a(b+2012)=b(a+2012)

<=>a/b=a+2012/b+2012

*a<b<=>2012a<2012b

<=>a(b+2012)<b(a+20120

<=>a/b<a+2012/b+2012

KL: a>b <=>a/b>a+2012/b+2012

....(tương tự như trên)
 

\(\frac{a}{b}=\frac{a\left(b+2012\right)}{b\left(b+2012\right)}=\frac{ab+2012a}{b\left(b+2012\right)}\)

\(\frac{a+2012}{b+2012}=\frac{\left(a+2012\right)b}{b\left(b+2012\right)}=\frac{ab+2012b}{b\left(b+2012\right)}\)

Vì b > 0 nên b(b + 2012) > 0 

a < 0 ; b > 0 nên a < b => 2012a < 2012b => ab + 2012a < ab + 2012b => \(\frac{ab+2012a}{b\left(b+2012\right)}

\(\frac{a}{b}< \frac{a+20}{b+20}\)

x^2+7x+2 chia hết cho x+7

x(x+7)+2 chia hết cho x+7

Vì x+7 chia hết cho x+7 nên x(x+7) chia hết cho x+7

=>2 chia hết cho x+7

hay x+7EƯ(2)={1;-1;2;-2}

=>xE{-6;-8;-5;-9}

Vậy để (x^2+7x+2) chia hết cho x+7 thì xE{-9;-8;-6;-5}