Question

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

Answer 1

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 \(x=13,5;y=\frac{-10}{3}\)

 

Answer 2

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 \(x=13,5;y=-5\)