Question

Tìm các giá trị của x, y thỏa mãn :

| 2x  - 27  |²⁰¹¹ + ( 3y + 10 )²⁰¹²=0

Answer 1

Ta thấy \(\left|2x-27\right|\ge0\Rightarrow\left|2x-27\right|^{2011}\ge0\)với mọi x

\(\left(3y+10\right)^{2012}\ge0\)với mọi y

Suy ra \( \left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)với mọi x,y

Mà \( \left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\)

Khi đó \(\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}}\)

Vậy.....

Answer 2

Ta có : \(\hept{\begin{cases}\left|2x-27\right|^{2011}\ge0\\\left(3y+10\right)^{2012}\ge0\end{cases}\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0}\)

Mà \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x=27\\3y=-10\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{27}{2}\\y=\frac{-10}{3}\end{cases}}}\)

Answer 3

|2x−27|2011+(3y+10)2012=0|2x−27|2011+(3y+10)2012=0

⇒|2x−27|2011=0⇒|2x−27|2011=0 và (3y+10)2012=0(3y+10)2012=0

+) |2x−27|2011=0|2x−27|2011=0

$\Rightarrow \left|2x-27\right|=0$

$\Rightarrow 2x-27=0$

$\Rightarrow 2x=27$

$\Rightarrow x=13,5$

+) (3y+10)2012=0(3y+10)2012=0

$\Rightarrow 3y+10=0$

$\Rightarrow 3y=-10$

$\Rightarrow y=\frac{-10}{3}$

Vậy $x=13,5;y=\frac{-10}{3}$