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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

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