a) ( a – 2009 ) ² + ( b + 2010 ) ² = 0 b) | a – 2010 | = 2009 15/08/2021 Bởi Margaret a) ( a – 2009 ) ² + ( b + 2010 ) ² = 0 b) | a – 2010 | = 2009
a) $(a-2009)^2≥0$ $(b+2010)^2≥0$ mà $(a-2009)^2+(b+2010)^2=0$ \(⇒\left[ \begin{array}{l}(a-2009)^2=0\\(b+2010)^2=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}a-2009=0\\b+2010=0\end{array} \right.\) \(⇒\left[ \begin{array}{l}a=2009\\b=-2010\end{array} \right.\) Vậy $(a,b)=(2009,-2010)$ b) $|a-2010|=2009$ \(⇒\left[ \begin{array}{l}a-2010=2009\\a-2010=-2009\end{array} \right.\) \(⇒\left[ \begin{array}{l}a=4019\\a=1\end{array} \right.\) Vậy $a=(4019,1)$ Bình luận
$a$) $(a-2009)^2 + (b+2010)^2 = 0$ Vì : $(a-2009)^2;(b+2010)^2 ≥ 0 ∀ a;b$ $⇒ a-2009=b+2010=0$ $⇒ a = 2009;b=-2010$ Vậy `(a;b);(2009;-2010)` $b$) `|a – 2010| = 2009` `⇔` \(\left[ \begin{array}{l}a=4019\\a=1\end{array} \right.\) Vậy $a$ $∈$ `{4019;1}` Bình luận
a) $(a-2009)^2≥0$
$(b+2010)^2≥0$
mà $(a-2009)^2+(b+2010)^2=0$
\(⇒\left[ \begin{array}{l}(a-2009)^2=0\\(b+2010)^2=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}a-2009=0\\b+2010=0\end{array} \right.\)
\(⇒\left[ \begin{array}{l}a=2009\\b=-2010\end{array} \right.\)
Vậy $(a,b)=(2009,-2010)$
b) $|a-2010|=2009$
\(⇒\left[ \begin{array}{l}a-2010=2009\\a-2010=-2009\end{array} \right.\)
\(⇒\left[ \begin{array}{l}a=4019\\a=1\end{array} \right.\)
Vậy $a=(4019,1)$
$a$) $(a-2009)^2 + (b+2010)^2 = 0$
Vì : $(a-2009)^2;(b+2010)^2 ≥ 0 ∀ a;b$
$⇒ a-2009=b+2010=0$
$⇒ a = 2009;b=-2010$
Vậy `(a;b);(2009;-2010)`
$b$) `|a – 2010| = 2009`
`⇔` \(\left[ \begin{array}{l}a=4019\\a=1\end{array} \right.\)
Vậy $a$ $∈$ `{4019;1}`