Bài $10$.
$d$) `|x-y-5| + 2003.(y-3)^{2014} = 0`
Ta có: `|x-y-5| ; 2003.(y-3)^{2014} ≥ 0` $∀$ $x;y$
$⇒$ `|x-y-5| + 2003.(y-3)^{2014} = 0` $⇔$ $\left\{\begin{matrix} |x-y-5|=0\\2013.(y-3)^{2014}=0 \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix} x-y=5\\y-3=0 \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix} x=5+y\\y=3 \end{matrix}\right.$
$⇔$ $\left\{\begin{matrix} x=8\\y=3 \end{matrix}\right.$
Vậy `(x;y)=(8;3)`.
Bài $12$.
$b$) `|x+1| + |x+2| + |x+3| + |x+4| = 5x-1`
Ta có: `|x+1| + |x+2| + |x+3| + |x+4| ≥ 0` $∀$ $x$
$⇒$ $5x-1 ≥ 0 ⇔ x ≥ \dfrac{1}{5} > 0$
$⇒$ `|x+1| + |x+2| + |x+3| + |x+4| = (x+1)+(x+2)+(x+3)+(x+4) = 5x-1`
$⇔ 4x + 10 = 5x-1$
$⇔ 5x-4x = 10 +1$
$⇔ x = 11$
Vậy $x=11$.
