Tìm x,y : a)|3x-4|+(y-2)²=0 b) (x-5)⁸+|y²-4|=0 c) |3x-4|+|3y+5|=0 d) |x-y|+|y+9/25|=0 e) (x+y)²⁰⁰⁶+2007|y-1|=0 f)|x-2007|+|y-2008|< hoặc =0 g)|x-y-2|+|y+3|=0 h)|x-3y|²⁰⁰⁷+|y+4|²⁰⁰⁸=0 m)|x-y-5|+2007(y-3)²⁰⁰⁸=0

$a$) `|3x-4| + (y-2)^2 = 0`

 Vì : `|3x-4|;(y-2)^2 ≥ 0` `∀` $x;y$

$⇒$ $3x-4=y-2=0$

$⇒$ $\left\{\begin{matrix}x =\dfrac{4}{3} & \\ y = 2 & \end{matrix}\right.$

   Vậy `(x;y)=(4/3;2)`

$b$) ` (x-5)^8+|y^2-4|=0`

 Vì : `(x-5)^8;|y^2-4|` `≥` `0` `∀` $x;y$

$⇒$ $x-5=y^2-4=0$

$⇒$ $\left\{\begin{matrix}x = 5 & \\ y = ±2 & \end{matrix}\right.$

    Vậy `(x;y)=(5;2);(5;-2)`

$c)$ `|3x-4|+|3y+5|=0`

Vì : $|3x-4|;|3y+5|$ $≥$ $0$ $∀$ $x;y$

$⇒ 3x-4=3y+5=0$

$⇒$ $\left\{\begin{matrix}x = \dfrac{4}{3} & \\ y = \dfrac{-5}{3} & \end{matrix}\right.$

   Vậy `(x;y)=(4/3;-5/3)`

$d$) `|x-y|+|y+9/{25}|=0`

Vì : $|x-y|;|y+\dfrac{9}{25}|$ $≥$ $0$ $∀$ $x;y$

$⇒ x-y=y+\dfrac{9}{25} = 0$

$⇒$ $\left\{\begin{matrix}x -y=0 & \\ y = -\dfrac{9}{25} & \end{matrix}\right.$

$⇒$ $\left\{\begin{matrix}x + \dfrac{9}{25}=0 & \\ y = -\dfrac{9}{25} & \end{matrix}\right.$

$⇒$  $\left\{\begin{matrix}x = - \dfrac{9}{25} & \\ y = -\dfrac{9}{25} & \end{matrix}\right.$

   Vậy `(x;y)=({-9}/{25};{-9}/{25})`

$e$) `(x+y)^{2006} + 2007.|y-1|=0`

Vì : $(x+y)^{2006};2017.|y-1|$ $≥$ $0$ $∀$ $x;y$

$⇒ x+y=2017.|y-1|=0$

$⇒$ $\left\{\begin{matrix}x + y=0 & \\ y = 1& \end{matrix}\right.$

$⇒$ $\left\{\begin{matrix}x = -1 & \\ y = 1& \end{matrix}\right.$

   Vậy `(x;y)=(-1;1)`

$f$) `|x-2007| + |y-2008| ≤ 0`

 Vì : `|x-2007|;|y-2008|≥0` `∀` $x;y$

$⇒$ $|x-2007|=|y-2008|=0$

$⇒$ $\left\{\begin{matrix}x =2007 & \\ y = 2008& \end{matrix}\right.$

   Vậy `(x;y)=(2007;2008)`

$g$) `|x-y-2|+|y+3|=0`

Vì : $|x-y-2|;|y+3|$ $≥0$ $∀$ $x;y$

$⇒ |x-y-2|=|y+3|=0$

$⇒$ $\left\{\begin{matrix}x - y - 2=0 & \\ y =-3& \end{matrix}\right.$

$⇒$ $\left\{\begin{matrix}x =-1 & \\ y =-3& \end{matrix}\right.$

   Vậy `(x;y)=(-1;-3)`

$h$) `|x-3y|^{2007} + |y+4|^{2008} = 0`

Vì : $|x-3y|^{2007};|y+4|^{2008}$ $≥$ $0$ $∀$ $x;y$

$⇒ |x-3y|=|y+4|=0$

$⇒$ $\left\{\begin{matrix}x - 3y=0 & \\ y =-4& \end{matrix}\right.$

$⇒$ $\left\{\begin{matrix}x = -12 & \\ y =-4& \end{matrix}\right.$

   Vậy `(x;y)=(-12;-4)`

$m$) `|x-y-5| + 2007.(y-3)^{2008} = 0`

 Vì : `|x-y-5|;2007.(y-3)^{2008} = 0`

`⇒ x-y-5=y-3=0`

`⇒` $\left\{\begin{matrix}x - y - 5 =0  & \\ y =3& \end{matrix}\right.$

$⇒$ $\left\{\begin{matrix}x = 8  & \\ y =3& \end{matrix}\right.$

   Vậy `(x;y)=(8;3)`