Vì \(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\text{≥0,∀x}\\\left(3y+10\right)^{2012}\text{≥0,∀y}\end{matrix}\right.\)

⇒ \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\text{≥0,∀x},y\)

Dấu "=" ⇔ \(\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy ...

Ta có \(\hept{\begin{cases}\left|2x-27\right|^{2011}\ge0\forall x\\\left(3y+10\right)^{2022}\ge0\forall y\end{cases}}\Rightarrow\left|2x-27\right|^{2011}+\left(3y+10\right)^{2022}\ge0\forall x;y\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)

Vậy x = 27/2 ; y = -10/3 là giá trị cần tìm

ta có |2x-27| > hoặc = 0=> |2x-27|^2011> hoặc = 0

(3y+10)^2012> hoặc 0 mà |2x-27|^2011+(3y+10)^2012=0 

=>2x-27=0 hoặc 3y+10=0=>2x=27 hoặc 3y=-10

=>x=13,5 hoặc x=-10/3

vậy .............................

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

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

Dấu ''='' xảy ra \(x=\frac{27}{2};y=-\frac{10}{3}\)

X=?

Y=?

Tìm các giá trị của x, y thỏa mãn: |2x-27|²⁰¹¹+(3y+10)²⁰¹²=0

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

Kết hợp với giả thiết ta thấy \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}=0\) nên:

\(\hept{\begin{cases}\left|2x-27\right|^{2011}=0\\\left(3y+10\right)^{2012}=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}2x-27=0\\3y+10=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}}\)

Vậy x=\(\frac{27}{2}\);y=\(-\frac{10}{3}\) thỏa mãn bài toán

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.....

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

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

|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

$\Rightarrow 3y+10=0$

Vì 

|2x - 27|²⁰¹¹ ≥ 0

(3y + 10)²⁰¹² ≥ 0

=> |2x - 27|²⁰¹¹ + (3y + 10)²⁰¹² ≥ 0

Dấu "=" xảy ra <=> |2x - 27|²⁰¹¹  = 0 và (3y + 10)²⁰¹² =0

<=> 2x - 27 = 0 và 3y + 10 = 0

=> x = 27/2 và y = - 10/3

ngu người

|2x - 27|²⁰¹¹ + (3y + 10)²⁰¹² = 0

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

\(\Rightarrow\begin{cases}\left|2x-27\right|=0\\3y+10=0\end{cases}\)

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

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

\(\Rightarrow\begin{cases}x=\frac{27}{2}\\y=-\frac{10}{3}\end{cases}\)

Vì \(\left|2x-27\right|\ge0\Rightarrow\left|2x-27\right|^{2011}\ge0\);  \(\left(3y+10\right)^{2012}\ge0\)

=>\(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\ge0\)

Dấu "=" xảy ra khi \(\left|2x-27\right|^{2011}=\left(3y+10\right)^{2012}=0\Leftrightarrow\hept{\begin{cases}\left|2x-27\right|=0\\\left(3y+10\right)^{2012}=0\end{cases}}\)

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

Vì \(\left\{{}\begin{matrix}\left|2x-27\right|^{2011}\text{≥0,∀x}\\\left(3y+10\right)^{2012}\text{≥0,∀y}\end{matrix}\right.\)

⇒ \(\left|2x-27\right|^{2011}+\left(3y+10\right)^{2012}\text{≥0,∀x},y\)

Dấu "=" ⇔ \(\left\{{}\begin{matrix}2x-27=0\\3y+10=0\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x=\dfrac{27}{2}\\y=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy ...