Question

[Cao Ngọc Hà]

 Ai giúp mình với

Answer 1

[Akai Haruma]

Bài 15:

a. Với $x$ nguyên,  để $A=\frac{x+5}{x-2}$ nguyên thì:

$x+5\vdots x-2$

$\Rightarrow x-2+7\vdots x-2$

$\Rightarrow 7\vdots x-2$

$\Rightarrow x-2\in \left\{\pm 1; \pm 7\right\}$

$\Rightarrow x\in \left\{3; 1; 9; -5\right\}$

b.

Với $x$ nguyên, để $A=\frac{2x-3}{x+1}$ nguyên thì:

$2x-3\vdots x+1$

$\Rightarrow 2(x+1)-5\vdots x+1$

$\Rightarrow 5\vdots x+1$
$\Rightarrow x+1\in \left\{\pm 1; \pm 5\right\}$

$\Rightarrow x\in \left\{0; -2; 4; -6\right\}$

Answer 2

[Akai Haruma]

Bài 16:

a. 

\(2009.\frac{2009^{2008}+1}{2009^{2009}+1}=\frac{2009^{2009}+2009}{2009^{2009}+1}=1+\frac{2008}{2009^{2009}+1}< 1+\frac{2008}{2009^{2008}+1}=\frac{2009^{2008}+2009}{2009^{2008}+1}=2009.\frac{2009^{2007}+1}{2009^{2008}+1}\)

\(\Rightarrow \frac{2009^{2008}+1}{2009^{2009}+1}< \frac{2009^{2007}+1}{2009^{2008}+1}\)

b.

\(\frac{7^{58}+2}{7^{57}+2}=\frac{7(7^{57}+2)-12}{7^{57}+2}=7-\frac{12}{7^{57}+2}\)

\(\frac{7^{57}+2009}{7^{56}+2009}=\frac{7(7^{56}+2009)-12054}{7^{56}+2009}=7-\frac{12054}{7^{56}+2009}\)

Ta thấy:

\(\frac{12}{7^{57}+2}=\frac{6}{\frac{7^{57}}{2}+1}<\frac{6}{\frac{7^{56}}{2009}+1}= \frac{12054}{7^{56}+2009}\)

\(\Rightarrow 7-\frac{12}{7^{57}+2}> 7-\frac{12054}{7^{56}+2009}\Rightarrow \frac{7^{58}+2}{7^{57}+2}>\frac{7^{57}+2009}{7^{56}+2009}\)