\(\frac{x+4}{2010}+\frac{x+3}{2011}=\frac{x+2}{2012}+\frac{x+1}{2013}\)

\(\Leftrightarrow\left(\frac{x+4}{2010}+1\right)+\left(\frac{x+3}{2011}+1\right)=\left(\frac{x+2}{2012}+1\right)+\left(\frac{x+1}{2013}+1\right)\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}=\frac{x+2014}{2012}+\frac{x+2014}{2013}\)

\(\Leftrightarrow\frac{x+2014}{2010}+\frac{x+2014}{2011}-\frac{x+2014}{2012}-\frac{x+2014}{2013}=0\)

\(\Leftrightarrow\left(x+2014\right)\left(\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}\right)=0\)

\(\Leftrightarrow x+2014=0\)

\(\Leftrightarrow x=-2014\)

V...

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 \(x=\frac{27}{2};y=-\frac{10}{3}\)

bạn ơi giải giúp mình nhé nhanh lên

Không tìm được

kho lam bn oi

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

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

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

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

\(\Rightarrow\left|2x-27\right|=0\)

\(\Rightarrow2x-27=0\)

\(\Rightarrow2x=27\)

\(\Rightarrow x=13,5\)

+) \(\left(3y+10\right)^{2012}=0\)

\(\Rightarrow3y+10=0\)

\(\Rightarrow3y=-10\)

\(\Rightarrow y=\frac{-10}{3}\)

Vậy \(x=13,5;y=\frac{-10}{3}\)

Ta có:

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

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

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

\(\Rightarrow\left|2x-27\right|=0\)

\(\Rightarrow2x-27=0\)

\(\Rightarrow2x=27\)

\(\Rightarrow x=13,5\)

+) \(\left(2y+10\right)^{2012}=0\)

\(\Rightarrow2y+10=0\)

\(\Rightarrow2y=-10\)

\(\Rightarrow y=-5\)

Vậy \(x=13,5;y=-5\)

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