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\(A=\frac{2n-1}{n+8}-\frac{n-14}{n+8}=\frac{2n-1-\left(n-14\right)}{n+8}=\frac{n+13}{n+8}\)
Để A thuộc Z thì \(n+13⋮n+8\Rightarrow n+13-\left(n+8\right)⋮n+8\)
\(\Rightarrow5⋮n+8\Rightarrow n+8\inƯ\left(5\right)=\left\{1;5;-1;-5\right\}\)
\(\Leftrightarrow n\in\left\{-7;-3;-9;-13\right\}\)
OK
\(\frac{2n+1}{n+3}=\frac{n+n+1}{n+3}=\frac{n}{n+3}+\frac{n+1}{n+3}\)
Do: \(\frac{n}{n+3}< \frac{n}{n+1};\frac{n+1}{n+3}< \frac{n+1}{n+2}\Rightarrow\frac{n}{n+3}+\frac{n+1}{n+3}< \frac{n}{n+1}+\frac{n+1}{n+2}\Rightarrow\frac{2n+1}{n+3}< \frac{n}{n+1}+\frac{n+1}{n+2}\)
a/
\(x-y=\frac{a}{b}-\frac{c}{d}=\frac{ad-cb}{bd}=\frac{1}{bd}.\) (1)
\(y-z=\frac{c}{d}-\frac{e}{h}=\frac{ch-de}{dh}=\frac{1}{dh}\)(2)
+ Nếu d>0 => (1)>0 và (2)>0 => x>y; y>x => x>y>z
+ Nếu d<0 => (1)<0 và (2)<0 => x<y; y<z => x<y<z
b/
\(m-y=\frac{a+e}{b+h}-\frac{c}{d}=\frac{ad+de-cb-ch}{d\left(b+h\right)}=\frac{\left(ad-cb\right)-\left(ch-de\right)}{d\left(b+h\right)}=\frac{1-1}{d\left(b+h\right)}=0\)
=> m=y
+
cảm ơn bn nha Nguyễn Ngoc Anh Minh mk k cho bn r đó kb vs mk nha
a: \(\Leftrightarrow3^n:27^n=\dfrac{1}{9}\)
\(\Leftrightarrow\left(\dfrac{1}{9}\right)^n=\dfrac{1}{9}\)
hay n=1
b: \(\Leftrightarrow3^n\cdot3^2=3^8\)
=>n+2=8
hay n=6
c: \(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot2^5\)
\(\Leftrightarrow2^n=2^6\)
hay n=6
d: \(\Leftrightarrow8^n=512\)
hay n=3
