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Theo bài ra ta có:
\(\frac{a}{b}=\frac{a+6}{b+9}\)
=> a(b+9)=b(a+6)
<=> ab+9a=ab+6b
<=> 9=6b
\(\Leftrightarrow\frac{a}{b}=\frac{6}{9}=\frac{2}{3}\)
Vậy \(\frac{a}{b}=\frac{2}{3}\)
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a)\(\left(x-\frac{1}{2}\right)^{2016},\left|\frac{3}{4}-y\right|\ge0\)
\(\left(x-\frac{1}{2}\right)^{2016}+\left|\frac{3}{4}-y\right|=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-\frac{1}{2}\right)^{2016}=0\\\left|\frac{3}{4}-y\right|=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=0\\\frac{3}{4}-y=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\y=\frac{3}{4}\end{cases}}\)
b)\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}\)
\(\Rightarrow\frac{b+c}{a}-\frac{a+c}{b}-\frac{a+b}{c}=0\)
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Khi a, b cùng dấu thì \(\frac{a}{b}>0\)
Khi a, b khác dấu thì \(\frac{a}{b}< 0\)