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\(\left(x+\frac{1}{2}\right)\left(x-\frac{3}{4}\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\x-\frac{3}{4}=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{4}\end{cases}}\)

\(A=\frac{2n-1}{n+8}-\frac{n-14}{n+8}=\frac{2n-1-\left(n-14\right)}{n+8}=\frac{n+13}{n+8}\)
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

Có: \(\frac{a}{b}=\frac{5}{6}=>\frac{a}{5}=\frac{b}{6}\)
Đặt \(\frac{a}{5}=\frac{b}{6}=k=>\hept{\begin{cases}a=5k\\b=6k\end{cases}}\)
Thay vào ta có:
A=\(\frac{3.5k-2.6k}{2.5k-3.5k}=\frac{15k-12k}{10k-15k}=\frac{3k}{-5k}=\frac{-3}{5}\)
=> \(A=\frac{-3}{5}\)

Ta có : \(A=\frac{2016^{2016}+2}{2016^{2016}-1}=\frac{2016^{2016}-1+3}{2016^{2016}-1}=1+\frac{3}{2016^{2016}-1}\)
\(B=\frac{2016^{2016}}{2016^{2016}-3}=\frac{2016^{2016}-3+3}{2016^{2016}-3}=1+\frac{3}{2016^{2016}-3}\)
Vì \(\frac{3}{2016^{2016}-1}>\frac{3}{2016^{2016}-3}\)
\(\Rightarrow1+\frac{3}{2016^{2016}-1}>1+\frac{3}{2016^{2016}-3}\)
\(\Rightarrow A>B\)

TH1: \(x\le-\frac{1}{2}\)
pt <=> \(\left[-\left(x-1\right)\right]-\left[-\left(2x+1\right)\right]=13\)<=>1-x+2x+1=13 <=> 2+x=13 <=> x=11 (loại)
TH2: \(-\frac{1}{2}< x\le1\)
pt <=> \(\left[-\left(x-1\right)\right]-\left(2x+1\right)=13\) <=> 1-x-2x-1=13 <=> -3x=13 <=> x=-13/3 (loại)
TH3: x > 1
pt <=> (x-1)-(2x+1)=13 <=> x-1-2x-1=13 <=> -x-2=13 <=> x=-15 (loại)
Vậy pt vô nghiệm

\(35\frac{1}{6}:\left(-\frac{4}{5}\right)-46\frac{1}{6}:\left(-\frac{4}{5}\right)\)
\(=\left(35\frac{1}{6}-46\frac{1}{6}\right):\left(-\frac{4}{5}\right)\)
\(=-11:\left(-\frac{4}{5}\right)\)
\(=\frac{55}{4}\)
211/6:(-4/5)-277/6:(-4/5)
(211/6-277/6):(-4/5)
11:(-4/5)
11.(-5/4)
-55/4
dau