Đáp án :
`a, (x;y) = (-5;4/3)`
`b, (x;y) = (1/2;1/4)`
`c, (x;y) = (-3;2)`
Giải thích các bước giải :
`a,`
`|x+5| + (3y-4)^{2012}=0`
Với mọi `x,y` có : \(\left\{ \begin{array}{l}|x+5|≥0\\(3y-4)^{2012}≥0\end{array} \right.\)
`⇔ |x+5| + (3y-4)^{2012} ≥0∀x,y`
Dấu "`=`" xảy ra khi :
`⇔` \(\left\{ \begin{array}{l}|x+5|=0\\(3y-4)^{2012}=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x+5=0\\3y-4=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=-5\\3y=4\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=-5\\y=\dfrac{4}{3}\end{array} \right.\)
Vậy `(x;y) = (-5;4/3)`
`b,`
`(2x-1)^2 + |2y-x| -8=12 -5×2^2`
`⇔ (2x-1)^2 + |2y-x|-8=12 - 5 ×4`
`⇔ (2x-1)^2 + |2y-x|-8=12-20`
`⇔ (2x-1)^2 + |2y-x|-8=-8`
`⇔ (2x-1)^2 + |2y-x|=0`
Với mọi `x,y` có : \(\left\{ \begin{array}{l}(2x-1)^2≥0\\|2y-x|≥0\end{array} \right.\)
`⇔ (2x-1)^2 + |2y-x|≥0∀x,y`
Dấu "`=`" xảy ra khi :
`⇔` \(\left\{ \begin{array}{l}(2x-1)^2=0\\|2y-x|=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}2x-1=0\\2y-x=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}2x=1\\2y=x\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=\dfrac{1}{2}\\2y=\dfrac{1}{2}\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=\dfrac{1}{2}\\y=\dfrac{1}{4}\end{array} \right.\)
Vậy `(x;y) = (1/2;1/4)`
`c,`
`(x+3)^{6034} + (2y-4)^{8912} ≤0`
Với mọi `x,y` có : \(\left\{ \begin{array}{l}(x+3)^{6034}≥0\\(2y-4)^{8912}≥0\end{array} \right.\)
`⇔ (x+3)^{6034} + (2y-4)^{8912} ≥0`
Dấu "`=`" xảy ra khi :
`⇔` \(\left\{ \begin{array}{l}(x+3)^{6034}=0\\(2y-4)^{8912}=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x+3=0\\2y-4=0\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=-3\\2y=4\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}x=-3\\y=2\end{array} \right.\)
Vậy `(x;y)=(-3;2)`
