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|+

0 bình luận về “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|+”

1. a)

   `|3x-4|+(y-2)^2≥0`

   Dấu = xảy ra

   `⇔3x-4=0;y-2=0`

   `⇔x=4/3;y=2`

   b)

   `(x-5)^8+|y^2-4|≥0`

   Dấu = xảy ra

   `⇔x-5=0;y^2-4=0`

   `⇔x=5;y=±2`

   c)

   ` |3x-4|+|3y+5|≥0`

   Dấu = xảy ra 

   `⇔3x-4=0;3y+5=0`

   `⇔x=4/3;y=-5/3`

   d)

   `|x-y|+|y+9/25|≥0`

   Dấu = xảy ra

   `⇔x-y=0;y+9/25`

   `⇔x=y=-9/25`

   e)

   `(x+y)^2006+2007|y-1|≥0`

   Dấu = xảy ra 

   `⇔x+y=0;y-1=0`

   `⇔x=-y;y=1`

   `⇔x=-1;y=1`

   f)

   `|x-2007|+|y-2008|≥0`

   `|x-2007|+|y-2008|≤0`

   `⇔|x-2007|+|y-2008|=0`

   `⇔x-2007=0;y-2008=0`

   `⇔x=2007;y=2008`

   g)

   `|x-y-2|+|y+3|≥0`

   Dấu = xảy ra

   `⇔x-y-2=0;y+3=0`

   `⇔x-y=2;y=-3`

   `⇔x=-1;y=-3`

   h)

   `|x-3y|^2007+|y+4|^2008≥0`

   Dấu = xảy ra

   `⇔x-3y=0;y+4=0`

   `⇔x=3y;y=-4`

   `⇔x=-12;y=-4`

   m)

   `|x-y-5|+2007(y-3)^2008≥0`

   Dấu = xảy ra

   `⇔x-y-5=0;y-3=0`

   `⇔x-y=5;y=3`

   `⇔x=8;y=3`

   Trả lời

2. $a$) `|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)`

   Trả lời