\(2x=5y\) và  \(y-x=12\)

\(2x=5y\Rightarrow\frac{x}{5}=\frac{y}{2}=\frac{y-x}{2-5}=\frac{12}{-3}=-4\)

\(+)\frac{x}{5}=-4\Rightarrow x=5.\left(-4\right)=-20\)

\(+)\frac{y}{2}=-4\Rightarrow y=2.\left(-4\right)=-8\)

Ta có : 

\(2x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{x}{5}=\frac{y}{2}=\frac{y-x}{2-5}=-\frac{12}{3}=\left(-4\right)\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=\left(-4\right)\Rightarrow x=-20\\\frac{y}{2}=\left(-4\right)\Rightarrow y=-8\end{cases}}\)

Vậy .... 

a) Vì \(\hept{\begin{cases}\left|5-x\right|\ge0,∀x\\\left(x-y\right)^2\ge0,∀x,y\end{cases}}\)=> A = 13 - | 5 - x | - ( x - y )² ≤ 0 

Dấu "=" xảy ra <=> \(\hept{\begin{cases}5-x=0\\x-y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=5\\x=y\end{cases}}\Leftrightarrow x=y=5\)

b) Ta có \(\orbr{\begin{cases}\left|x-5\right|=x-5\left(\text{nếu }x-5>0\right)\\\left|x-5\right|=\left|5-x\right|=x-5\left(\text{nếu }x-5< 0\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}B\ge-\left(x-5\right)+x+4\\B\ge-\left(x-5\right)+x+4\end{cases}}\Rightarrow\orbr{\begin{cases}B\ge9\\B\ge9\end{cases}}\)<=> B ≥ 9

Dấu "=" xảy ra <=> x - 5 = 0 <=> x = 5

Ta có \(\hept{\begin{cases}\left(\frac{3x-5}{9}\right)^{2006}\ge0\forall x\\\left(\frac{3y+0,4}{3}\right)^{2008}\ge0\forall y\end{cases}}\Rightarrow\left(\frac{3x-5}{9}\right)^{2006}+\left(\frac{3y+0,4}{3}\right)^{2008}\ge0\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\frac{3x-5}{9}=0\\\frac{3y+0,4}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{2}{15}\end{cases}}\)

Vậy x = 5/3 ; y =- 2/15

\(a)\frac{1}{6}+\frac{1}{2}+\frac{1}{3}-1\)

\(=\frac{1}{6}+\frac{3}{6}+\frac{2}{6}-\frac{6}{6}\)

\(=\frac{0}{6}=0\)

b) \(\left(\frac{-1}{2}\right)^3+\left|\frac{-7}{6}\right|-\frac{7}{8}\)

\(=\frac{-1}{2^3}+\frac{7}{6}-\frac{7}{8}\)

\(=\frac{-1}{8}+\frac{7}{6}-\frac{7}{8}\)

\(=\left(\frac{-1}{8}-\frac{7}{8}\right)+\frac{7}{6}\)

\(=-1+\frac{7}{6}=\frac{1}{6}\)

c) \(\left(\frac{5}{4}-\frac{2}{5}\right).\frac{2015}{2016}+\left(\frac{3}{4}-\frac{3}{5}\right).\frac{2015}{2016}\)

\(=\left[\left(\frac{5}{4}-\frac{2}{5}\right)+\left(\frac{3}{4}-\frac{3}{5}\right)\right].\frac{2015}{2016}\)

\(=\left[\left(\frac{5}{4}+\frac{3}{4}\right)-\left(\frac{2}{5}+\frac{3}{5}\right)\right].\frac{2015}{2016}\)

\(=\left(1-1\right).\frac{2015}{2016}=0.\frac{2015}{2016}=0\)

d) \(\frac{\left(-5\right)^2.20^2}{8^2.125}\)\(=\frac{5^2.\left(2^2.5\right)^2}{\left(2^3\right)^2.5^3}=\frac{5^2.\left(2^2\right)^2.5^2}{2^6.5^3}=\frac{5^4.2^4}{2^6.5^3}=\frac{5}{2^2}=\frac{5}{4}\)